Solve x*3/4*x=96

$$x \times \frac{ 3 }{ 4 } \times x=96$$

Solve for x

$$x=8\sqrt{2}\approx 11.313708499$$

$$x=-8\sqrt{2}\approx -11.313708499$$

Steps by Finding Square Root

Multiply x and x to get x^{2}.

$$x^{2}\times \frac{3}{4}=96$$

Multiply both sides by \frac{4}{3}, the reciprocal of \frac{3}{4}.

$$x^{2}=96\times \frac{4}{3}$$

Express 96\times \frac{4}{3} as a single fraction.

$$x^{2}=\frac{96\times 4}{3}$$

Multiply 96 and 4 to get 384.

$$x^{2}=\frac{384}{3}$$

Divide 384 by 3 to get 128.

$$x^{2}=128$$

Take the square root of both sides of the equation.

$$x=8\sqrt{2} x=-8\sqrt{2}$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$x^{2}\times \frac{3}{4}=96$$

Subtract 96 from both sides.

$$x^{2}\times \frac{3}{4}-96=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$\frac{3}{4}x^{2}-96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{4} for a, 0 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{3}{4}\left(-96\right)}}{2\times \frac{3}{4}}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times \frac{3}{4}\left(-96\right)}}{2\times \frac{3}{4}}$$

Multiply -4 times \frac{3}{4}.

$$x=\frac{0±\sqrt{-3\left(-96\right)}}{2\times \frac{3}{4}}$$

Multiply -3 times -96.

$$x=\frac{0±\sqrt{288}}{2\times \frac{3}{4}}$$

Take the square root of 288.

$$x=\frac{0±12\sqrt{2}}{2\times \frac{3}{4}}$$

Multiply 2 times \frac{3}{4}.

$$x=\frac{0±12\sqrt{2}}{\frac{3}{2}}$$

Now solve the equation x=\frac{0±12\sqrt{2}}{\frac{3}{2}} when ± is plus.

$$x=8\sqrt{2}$$

Now solve the equation x=\frac{0±12\sqrt{2}}{\frac{3}{2}} when ± is minus.

$$x=-8\sqrt{2}$$

The equation is now solved.

$$x=8\sqrt{2} x=-8\sqrt{2}$$