Solve 5 2y^ +2 +3^ +3 +; y^ +2 -5^ ^ prime2 x+2; 4 + 3 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac { 2 y ^ { 2 } + 3 + 3 } { 4 + 3 } + \frac { y + 25 } { 6 + 2 } = 0$$

Solve for y

$y=\frac{-7+\sqrt{14223}i}{32}\approx -0.21875+3.726881865i$
$y=\frac{-\sqrt{14223}i-7}{32}\approx -0.21875-3.726881865i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Add $3$ and $3$ to get $6$.

$$\frac{2y^{2}+6}{4+3}+\frac{y+25}{6+2}=0$$

Add $4$ and $3$ to get $7$.

$$\frac{2y^{2}+6}{7}+\frac{y+25}{6+2}=0$$

Add $6$ and $2$ to get $8$.

$$\frac{2y^{2}+6}{7}+\frac{y+25}{8}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $7$ and $8$ is $56$. Multiply $\frac{2y^{2}+6}{7}$ times $\frac{8}{8}$. Multiply $\frac{y+25}{8}$ times $\frac{7}{7}$.

$$\frac{8\left(2y^{2}+6\right)}{56}+\frac{7\left(y+25\right)}{56}=0$$

Since $\frac{8\left(2y^{2}+6\right)}{56}$ and $\frac{7\left(y+25\right)}{56}$ have the same denominator, add them by adding their numerators.

$$\frac{8\left(2y^{2}+6\right)+7\left(y+25\right)}{56}=0$$

Do the multiplications in $8\left(2y^{2}+6\right)+7\left(y+25\right)$.

$$\frac{16y^{2}+48+7y+175}{56}=0$$

Combine like terms in $16y^{2}+48+7y+175$.

$$\frac{16y^{2}+223+7y}{56}=0$$

Divide each term of $16y^{2}+223+7y$ by $56$ to get $\frac{2}{7}y^{2}+\frac{223}{56}+\frac{1}{8}y$.

$$\frac{2}{7}y^{2}+\frac{223}{56}+\frac{1}{8}y=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$\frac{2}{7}y^{2}+\frac{1}{8}y+\frac{223}{56}=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $\frac{2}{7}$ for $a$, $\frac{1}{8}$ for $b$, and $\frac{223}{56}$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-\frac{1}{8}±\sqrt{\left(\frac{1}{8}\right)^{2}-4\times \frac{2}{7}\times \frac{223}{56}}}{2\times \frac{2}{7}}$$

Square $\frac{1}{8}$ by squaring both the numerator and the denominator of the fraction.

$$y=\frac{-\frac{1}{8}±\sqrt{\frac{1}{64}-4\times \frac{2}{7}\times \frac{223}{56}}}{2\times \frac{2}{7}}$$

Multiply $-4$ times $\frac{2}{7}$.

$$y=\frac{-\frac{1}{8}±\sqrt{\frac{1}{64}-\frac{8}{7}\times \frac{223}{56}}}{2\times \frac{2}{7}}$$

Multiply $-\frac{8}{7}$ times $\frac{223}{56}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{1}{8}±\sqrt{\frac{1}{64}-\frac{223}{49}}}{2\times \frac{2}{7}}$$

Add $\frac{1}{64}$ to $-\frac{223}{49}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{1}{8}±\sqrt{-\frac{14223}{3136}}}{2\times \frac{2}{7}}$$

Take the square root of $-\frac{14223}{3136}$.

$$y=\frac{-\frac{1}{8}±\frac{\sqrt{14223}i}{56}}{2\times \frac{2}{7}}$$

Multiply $2$ times $\frac{2}{7}$.

$$y=\frac{-\frac{1}{8}±\frac{\sqrt{14223}i}{56}}{\frac{4}{7}}$$

Now solve the equation $y=\frac{-\frac{1}{8}±\frac{\sqrt{14223}i}{56}}{\frac{4}{7}}$ when $±$ is plus. Add $-\frac{1}{8}$ to $\frac{i\sqrt{14223}}{56}$.

$$y=\frac{\frac{\sqrt{14223}i}{56}-\frac{1}{8}}{\frac{4}{7}}$$

Divide $-\frac{1}{8}+\frac{i\sqrt{14223}}{56}$ by $\frac{4}{7}$ by multiplying $-\frac{1}{8}+\frac{i\sqrt{14223}}{56}$ by the reciprocal of $\frac{4}{7}$.

$$y=\frac{-7+\sqrt{14223}i}{32}$$

Now solve the equation $y=\frac{-\frac{1}{8}±\frac{\sqrt{14223}i}{56}}{\frac{4}{7}}$ when $±$ is minus. Subtract $\frac{i\sqrt{14223}}{56}$ from $-\frac{1}{8}$.

$$y=\frac{-\frac{\sqrt{14223}i}{56}-\frac{1}{8}}{\frac{4}{7}}$$

Divide $-\frac{1}{8}-\frac{i\sqrt{14223}}{56}$ by $\frac{4}{7}$ by multiplying $-\frac{1}{8}-\frac{i\sqrt{14223}}{56}$ by the reciprocal of $\frac{4}{7}$.

$$y=\frac{-\sqrt{14223}i-7}{32}$$

The equation is now solved.

$$y=\frac{-7+\sqrt{14223}i}{32}$$ $$y=\frac{-\sqrt{14223}i-7}{32}$$

Steps for Completing the Square

Add $3$ and $3$ to get $6$.

$$\frac{2y^{2}+6}{4+3}+\frac{y+25}{6+2}=0$$

Add $4$ and $3$ to get $7$.

$$\frac{2y^{2}+6}{7}+\frac{y+25}{6+2}=0$$

Add $6$ and $2$ to get $8$.

$$\frac{2y^{2}+6}{7}+\frac{y+25}{8}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $7$ and $8$ is $56$. Multiply $\frac{2y^{2}+6}{7}$ times $\frac{8}{8}$. Multiply $\frac{y+25}{8}$ times $\frac{7}{7}$.

$$\frac{8\left(2y^{2}+6\right)}{56}+\frac{7\left(y+25\right)}{56}=0$$

Since $\frac{8\left(2y^{2}+6\right)}{56}$ and $\frac{7\left(y+25\right)}{56}$ have the same denominator, add them by adding their numerators.

$$\frac{8\left(2y^{2}+6\right)+7\left(y+25\right)}{56}=0$$

Do the multiplications in $8\left(2y^{2}+6\right)+7\left(y+25\right)$.

$$\frac{16y^{2}+48+7y+175}{56}=0$$

Combine like terms in $16y^{2}+48+7y+175$.

$$\frac{16y^{2}+223+7y}{56}=0$$

Divide each term of $16y^{2}+223+7y$ by $56$ to get $\frac{2}{7}y^{2}+\frac{223}{56}+\frac{1}{8}y$.

$$\frac{2}{7}y^{2}+\frac{223}{56}+\frac{1}{8}y=0$$

Subtract $\frac{223}{56}$ from both sides. Anything subtracted from zero gives its negation.

$$\frac{2}{7}y^{2}+\frac{1}{8}y=-\frac{223}{56}$$

Divide both sides of the equation by $\frac{2}{7}$, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{2}{7}y^{2}+\frac{1}{8}y}{\frac{2}{7}}=-\frac{\frac{223}{56}}{\frac{2}{7}}$$

Dividing by $\frac{2}{7}$ undoes the multiplication by $\frac{2}{7}$.

$$y^{2}+\frac{\frac{1}{8}}{\frac{2}{7}}y=-\frac{\frac{223}{56}}{\frac{2}{7}}$$

Divide $\frac{1}{8}$ by $\frac{2}{7}$ by multiplying $\frac{1}{8}$ by the reciprocal of $\frac{2}{7}$.

$$y^{2}+\frac{7}{16}y=-\frac{\frac{223}{56}}{\frac{2}{7}}$$

Divide $-\frac{223}{56}$ by $\frac{2}{7}$ by multiplying $-\frac{223}{56}$ by the reciprocal of $\frac{2}{7}$.

$$y^{2}+\frac{7}{16}y=-\frac{223}{16}$$

Divide $\frac{7}{16}$, the coefficient of the $x$ term, by $2$ to get $\frac{7}{32}$. Then add the square of $\frac{7}{32}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{7}{16}y+\left(\frac{7}{32}\right)^{2}=-\frac{223}{16}+\left(\frac{7}{32}\right)^{2}$$

Square $\frac{7}{32}$ by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{7}{16}y+\frac{49}{1024}=-\frac{223}{16}+\frac{49}{1024}$$

Add $-\frac{223}{16}$ to $\frac{49}{1024}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{7}{16}y+\frac{49}{1024}=-\frac{14223}{1024}$$

Factor $y^{2}+\frac{7}{16}y+\frac{49}{1024}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(y+\frac{7}{32}\right)^{2}=-\frac{14223}{1024}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{7}{32}\right)^{2}}=\sqrt{-\frac{14223}{1024}}$$

Simplify.

$$y+\frac{7}{32}=\frac{\sqrt{14223}i}{32}$$ $$y+\frac{7}{32}=-\frac{\sqrt{14223}i}{32}$$

Subtract $\frac{7}{32}$ from both sides of the equation.

$$y=\frac{-7+\sqrt{14223}i}{32}$$ $$y=\frac{-\sqrt{14223}i-7}{32}$$