Solve -5x^2-61x-114=96+4x

$$- 5 x ^ { 2 } - 61 x - 114 = 96 + 4 x$$

Solve for x

$$x=-7$$

$$x=-6$$

Steps Using the Quadratic Formula

Subtract 96 from both sides.

$$-5x^{2}-61x-114-96=4x$$

Subtract 96 from -114 to get -210.

$$-5x^{2}-61x-210=4x$$

Subtract 4x from both sides.

$$-5x^{2}-61x-210-4x=0$$

Combine -61x and -4x to get -65x.

$$-5x^{2}-65x-210=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -65 for b, and -210 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-65\right)±\sqrt{\left(-65\right)^{2}-4\left(-5\right)\left(-210\right)}}{2\left(-5\right)}$$

Square -65.

$$x=\frac{-\left(-65\right)±\sqrt{4225-4\left(-5\right)\left(-210\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-65\right)±\sqrt{4225+20\left(-210\right)}}{2\left(-5\right)}$$

Multiply 20 times -210.

$$x=\frac{-\left(-65\right)±\sqrt{4225-4200}}{2\left(-5\right)}$$

Add 4225 to -4200.

$$x=\frac{-\left(-65\right)±\sqrt{25}}{2\left(-5\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-65\right)±5}{2\left(-5\right)}$$

The opposite of -65 is 65.

$$x=\frac{65±5}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{65±5}{-10}$$

Now solve the equation x=\frac{65±5}{-10} when ± is plus. Add 65 to 5.

$$x=\frac{70}{-10}$$

Divide 70 by -10.

$$x=-7$$

Now solve the equation x=\frac{65±5}{-10} when ± is minus. Subtract 5 from 65.

$$x=\frac{60}{-10}$$

Divide 60 by -10.

$$x=-6$$

The equation is now solved.

$$x=-7 x=-6$$

Steps for Completing the Square

Subtract 4x from both sides.

$$-5x^{2}-61x-114-4x=96$$

Combine -61x and -4x to get -65x.

$$-5x^{2}-65x-114=96$$

Add 114 to both sides.

$$-5x^{2}-65x=96+114$$

Add 96 and 114 to get 210.

$$-5x^{2}-65x=210$$

Divide both sides by -5.

$$\frac{-5x^{2}-65x}{-5}=\frac{210}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{65}{-5}\right)x=\frac{210}{-5}$$

Divide -65 by -5.

$$x^{2}+13x=\frac{210}{-5}$$

Divide 210 by -5.

$$x^{2}+13x=-42$$

Divide 13, the coefficient of the x term, by 2 to get \frac{13}{2}. Then add the square of \frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+13x+\left(\frac{13}{2}\right)^{2}=-42+\left(\frac{13}{2}\right)^{2}$$

Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+13x+\frac{169}{4}=-42+\frac{169}{4}$$

Add -42 to \frac{169}{4}.

$$x^{2}+13x+\frac{169}{4}=\frac{1}{4}$$

Factor x^{2}+13x+\frac{169}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{13}{2}=\frac{1}{2} x+\frac{13}{2}=-\frac{1}{2}$$

Subtract \frac{13}{2} from both sides of the equation.

$$x=-6 x=-7$$