Solve 8\times2y+2\times2y/-2y+1+4\times2y=5y | 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

$$8 \times 2y+2 \times 2y/-2y+1+4 \times 2y=5y$$

Solve for y

$y = \frac{3 \sqrt{41} + 19}{4} \approx 9.552343178$
$y=\frac{19-3\sqrt{41}}{4}\approx -0.052343178$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $-2$.

$$-16\times 2y+2\times 2yy-2-8\times 2y=-10y$$

Multiply $-16$ and $2$ to get $-32$. Multiply $-8$ and $2$ to get $-16$.

$$-32y+2\times 2yy-2-16y=-10y$$

Multiply $y$ and $y$ to get $y^{2}$.

$$-32y+2\times 2y^{2}-2-16y=-10y$$

Multiply $2$ and $2$ to get $4$.

$$-32y+4y^{2}-2-16y=-10y$$

Combine $-32y$ and $-16y$ to get $-48y$.

$$-48y+4y^{2}-2=-10y$$

Add $10y$ to both sides.

$$-48y+4y^{2}-2+10y=0$$

Combine $-48y$ and $10y$ to get $-38y$.

$$-38y+4y^{2}-2=0$$

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.

$$4y^{2}-38y-2=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $4$ for $a$, $-38$ for $b$, and $-2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-\left(-38\right)±\sqrt{\left(-38\right)^{2}-4\times 4\left(-2\right)}}{2\times 4}$$

Square $-38$.

$$y=\frac{-\left(-38\right)±\sqrt{1444-4\times 4\left(-2\right)}}{2\times 4}$$

Multiply $-4$ times $4$.

$$y=\frac{-\left(-38\right)±\sqrt{1444-16\left(-2\right)}}{2\times 4}$$

Multiply $-16$ times $-2$.

$$y=\frac{-\left(-38\right)±\sqrt{1444+32}}{2\times 4}$$

Add $1444$ to $32$.

$$y=\frac{-\left(-38\right)±\sqrt{1476}}{2\times 4}$$

Take the square root of $1476$.

$$y=\frac{-\left(-38\right)±6\sqrt{41}}{2\times 4}$$

The opposite of $-38$ is $38$.

$$y=\frac{38±6\sqrt{41}}{2\times 4}$$

Multiply $2$ times $4$.

$$y=\frac{38±6\sqrt{41}}{8}$$

Now solve the equation $y=\frac{38±6\sqrt{41}}{8}$ when $±$ is plus. Add $38$ to $6\sqrt{41}$.

$$y=\frac{6\sqrt{41}+38}{8}$$

Divide $38+6\sqrt{41}$ by $8$.

$$y=\frac{3\sqrt{41}+19}{4}$$

Now solve the equation $y=\frac{38±6\sqrt{41}}{8}$ when $±$ is minus. Subtract $6\sqrt{41}$ from $38$.

$$y=\frac{38-6\sqrt{41}}{8}$$

Divide $38-6\sqrt{41}$ by $8$.

$$y=\frac{19-3\sqrt{41}}{4}$$

The equation is now solved.

$$y=\frac{3\sqrt{41}+19}{4}$$ $$y=\frac{19-3\sqrt{41}}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by $-2$.

$$-16\times 2y+2\times 2yy-2-8\times 2y=-10y$$

Multiply $-16$ and $2$ to get $-32$. Multiply $-8$ and $2$ to get $-16$.

$$-32y+2\times 2yy-2-16y=-10y$$

Multiply $y$ and $y$ to get $y^{2}$.

$$-32y+2\times 2y^{2}-2-16y=-10y$$

Multiply $2$ and $2$ to get $4$.

$$-32y+4y^{2}-2-16y=-10y$$

Combine $-32y$ and $-16y$ to get $-48y$.

$$-48y+4y^{2}-2=-10y$$

Add $10y$ to both sides.

$$-48y+4y^{2}-2+10y=0$$

Combine $-48y$ and $10y$ to get $-38y$.

$$-38y+4y^{2}-2=0$$

Add $2$ to both sides. Anything plus zero gives itself.

$$-38y+4y^{2}=2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$4y^{2}-38y=2$$

Divide both sides by $4$.

$$\frac{4y^{2}-38y}{4}=\frac{2}{4}$$

Dividing by $4$ undoes the multiplication by $4$.

$$y^{2}+\left(-\frac{38}{4}\right)y=\frac{2}{4}$$

Reduce the fraction $\frac{-38}{4}$ to lowest terms by extracting and canceling out $2$.

$$y^{2}-\frac{19}{2}y=\frac{2}{4}$$

Reduce the fraction $\frac{2}{4}$ to lowest terms by extracting and canceling out $2$.

$$y^{2}-\frac{19}{2}y=\frac{1}{2}$$

Divide $-\frac{19}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{19}{4}$. Then add the square of $-\frac{19}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{19}{2}y+\left(-\frac{19}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{19}{4}\right)^{2}$$

Square $-\frac{19}{4}$ by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{19}{2}y+\frac{361}{16}=\frac{1}{2}+\frac{361}{16}$$

Add $\frac{1}{2}$ to $\frac{361}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{19}{2}y+\frac{361}{16}=\frac{369}{16}$$

Factor $y^{2}-\frac{19}{2}y+\frac{361}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(y-\frac{19}{4}\right)^{2}=\frac{369}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{19}{4}\right)^{2}}=\sqrt{\frac{369}{16}}$$

Simplify.

$$y-\frac{19}{4}=\frac{3\sqrt{41}}{4}$$ $$y-\frac{19}{4}=-\frac{3\sqrt{41}}{4}$$

Add $\frac{19}{4}$ to both sides of the equation.

$$y=\frac{3\sqrt{41}+19}{4}$$ $$y=\frac{19-3\sqrt{41}}{4}$$