Solve 96x=9x^2-144

$$96 x = 9 x ^ { 2 } - 144$$

Solve for x

$$x=12$$

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

Steps Using Factoring By Grouping

Subtract 9x^{2} from both sides.

$$96x-9x^{2}=-144$$

Add 144 to both sides.

$$96x-9x^{2}+144=0$$

Divide both sides by 3.

$$32x-3x^{2}+48=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}+32x+48=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+48. To find a and b, set up a system to be solved.

$$a+b=32 ab=-3\times 48=-144$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -144.

$$-1,144 -2,72 -3,48 -4,36 -6,24 -8,18 -9,16 -12,12$$

Calculate the sum for each pair.

$$-1+144=143 -2+72=70 -3+48=45 -4+36=32 -6+24=18 -8+18=10 -9+16=7 -12+12=0$$

The solution is the pair that gives sum 32.

$$a=36 b=-4$$

Rewrite -3x^{2}+32x+48 as \left(-3x^{2}+36x\right)+\left(-4x+48\right).

$$\left(-3x^{2}+36x\right)+\left(-4x+48\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(-x+12\right)+4\left(-x+12\right)$$

Factor out common term -x+12 by using distributive property.

$$\left(-x+12\right)\left(3x+4\right)$$

To find equation solutions, solve -x+12=0 and 3x+4=0.

$$x=12 x=-\frac{4}{3}$$

Steps Using the Quadratic Formula

Subtract 9x^{2} from both sides.

$$96x-9x^{2}=-144$$

Add 144 to both sides.

$$96x-9x^{2}+144=0$$

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.

$$-9x^{2}+96x+144=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 96 for b, and 144 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-96±\sqrt{96^{2}-4\left(-9\right)\times 144}}{2\left(-9\right)}$$

Square 96.

$$x=\frac{-96±\sqrt{9216-4\left(-9\right)\times 144}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-96±\sqrt{9216+36\times 144}}{2\left(-9\right)}$$

Multiply 36 times 144.

$$x=\frac{-96±\sqrt{9216+5184}}{2\left(-9\right)}$$

Add 9216 to 5184.

$$x=\frac{-96±\sqrt{14400}}{2\left(-9\right)}$$

Take the square root of 14400.

$$x=\frac{-96±120}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-96±120}{-18}$$

Now solve the equation x=\frac{-96±120}{-18} when ± is plus. Add -96 to 120.

$$x=\frac{24}{-18}$$

Reduce the fraction \frac{24}{-18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{4}{3}$$

Now solve the equation x=\frac{-96±120}{-18} when ± is minus. Subtract 120 from -96.

$$x=-\frac{216}{-18}$$

Divide -216 by -18.

$$x=12$$

The equation is now solved.

$$x=-\frac{4}{3} x=12$$

Steps for Completing the Square

Subtract 9x^{2} from both sides.

$$96x-9x^{2}=-144$$

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.

$$-9x^{2}+96x=-144$$

Divide both sides by -9.

$$\frac{-9x^{2}+96x}{-9}=-\frac{144}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{96}{-9}x=-\frac{144}{-9}$$

Reduce the fraction \frac{96}{-9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{32}{3}x=-\frac{144}{-9}$$

Divide -144 by -9.

$$x^{2}-\frac{32}{3}x=16$$

Divide -\frac{32}{3}, the coefficient of the x term, by 2 to get -\frac{16}{3}. Then add the square of -\frac{16}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{3}x+\left(-\frac{16}{3}\right)^{2}=16+\left(-\frac{16}{3}\right)^{2}$$

Square -\frac{16}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=16+\frac{256}{9}$$

Add 16 to \frac{256}{9}.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=\frac{400}{9}$$

Factor x^{2}-\frac{32}{3}x+\frac{256}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{16}{3}\right)^{2}=\frac{400}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{3}\right)^{2}}=\sqrt{\frac{400}{9}}$$

Simplify.

$$x-\frac{16}{3}=\frac{20}{3} x-\frac{16}{3}=-\frac{20}{3}$$

Add \frac{16}{3} to both sides of the equation.

$$x=12 x=-\frac{4}{3}$$