Solve 2x ^ 2 - 3x = 7 + 5x ^ 2 + 8x | Equatix - Math Solver

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Example

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

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Question

$$2x^{2}-3x=7+5x^{2}+8x$$

Solve for x

$x=\frac{\sqrt{37}-11}{6}\approx -0.819539578$
$x=\frac{-\sqrt{37}-11}{6}\approx -2.847127088$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $7$ from both sides.

$$2x^{2}-3x-7=5x^{2}+8x$$

Subtract $5x^{2}$ from both sides.

$$2x^{2}-3x-7-5x^{2}=8x$$

Combine $2x^{2}$ and $-5x^{2}$ to get $-3x^{2}$.

$$-3x^{2}-3x-7=8x$$

Subtract $8x$ from both sides.

$$-3x^{2}-3x-7-8x=0$$

Combine $-3x$ and $-8x$ to get $-11x$.

$$-3x^{2}-11x-7=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-3$ for $a$, $-11$ for $b$, and $-7$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-3\right)\left(-7\right)}}{2\left(-3\right)}$$

Square $-11$.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-3\right)\left(-7\right)}}{2\left(-3\right)}$$

Multiply $-4$ times $-3$.

$$x=\frac{-\left(-11\right)±\sqrt{121+12\left(-7\right)}}{2\left(-3\right)}$$

Multiply $12$ times $-7$.

$$x=\frac{-\left(-11\right)±\sqrt{121-84}}{2\left(-3\right)}$$

Add $121$ to $-84$.

$$x=\frac{-\left(-11\right)±\sqrt{37}}{2\left(-3\right)}$$

The opposite of $-11$ is $11$.

$$x=\frac{11±\sqrt{37}}{2\left(-3\right)}$$

Multiply $2$ times $-3$.

$$x=\frac{11±\sqrt{37}}{-6}$$

Now solve the equation $x=\frac{11±\sqrt{37}}{-6}$ when $±$ is plus. Add $11$ to $\sqrt{37}$.

$$x=\frac{\sqrt{37}+11}{-6}$$

Divide $11+\sqrt{37}$ by $-6$.

$$x=\frac{-\sqrt{37}-11}{6}$$

Now solve the equation $x=\frac{11±\sqrt{37}}{-6}$ when $±$ is minus. Subtract $\sqrt{37}$ from $11$.

$$x=\frac{11-\sqrt{37}}{-6}$$

Divide $11-\sqrt{37}$ by $-6$.

$$x=\frac{\sqrt{37}-11}{6}$$

The equation is now solved.

$$x=\frac{-\sqrt{37}-11}{6}$$ $$x=\frac{\sqrt{37}-11}{6}$$

Steps for Completing the Square

Subtract $5x^{2}$ from both sides.

$$2x^{2}-3x-5x^{2}=7+8x$$

Combine $2x^{2}$ and $-5x^{2}$ to get $-3x^{2}$.

$$-3x^{2}-3x=7+8x$$

Subtract $8x$ from both sides.

$$-3x^{2}-3x-8x=7$$

Combine $-3x$ and $-8x$ to get $-11x$.

$$-3x^{2}-11x=7$$

Divide both sides by $-3$.

$$\frac{-3x^{2}-11x}{-3}=\frac{7}{-3}$$

Dividing by $-3$ undoes the multiplication by $-3$.

$$x^{2}+\left(-\frac{11}{-3}\right)x=\frac{7}{-3}$$

Divide $-11$ by $-3$.

$$x^{2}+\frac{11}{3}x=\frac{7}{-3}$$

Divide $7$ by $-3$.

$$x^{2}+\frac{11}{3}x=-\frac{7}{3}$$

Divide $\frac{11}{3}$, the coefficient of the $x$ term, by $2$ to get $\frac{11}{6}$. Then add the square of $\frac{11}{6}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{3}x+\left(\frac{11}{6}\right)^{2}=-\frac{7}{3}+\left(\frac{11}{6}\right)^{2}$$

Square $\frac{11}{6}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=-\frac{7}{3}+\frac{121}{36}$$

Add $-\frac{7}{3}$ to $\frac{121}{36}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=\frac{37}{36}$$

Factor $x^{2}+\frac{11}{3}x+\frac{121}{36}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{11}{6}\right)^{2}=\frac{37}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{6}\right)^{2}}=\sqrt{\frac{37}{36}}$$

Simplify.

$$x+\frac{11}{6}=\frac{\sqrt{37}}{6}$$ $$x+\frac{11}{6}=-\frac{\sqrt{37}}{6}$$

Subtract $\frac{11}{6}$ from both sides of the equation.

$$x=\frac{\sqrt{37}-11}{6}$$ $$x=\frac{-\sqrt{37}-11}{6}$$