Solve 3/4x^2-7=5

$$3 / 4 x ^ { 2 } - 7 = 5$$

Solve for x

$$x=4$$

$$x=-4$$

Steps Using Difference of Squares

Subtract 5 from both sides.

$$\frac{3}{4}x^{2}-7-5=0$$

Subtract 5 from -7 to get -12.

$$\frac{3}{4}x^{2}-12=0$$

Divide both sides by \frac{3}{4}.

$$x^{2}-16=0$$

Consider x^{2}-16. Rewrite x^{2}-16 as x^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)=0$$

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

$$x=4 x=-4$$

Steps by Finding Square Root

Add 7 to both sides.

$$\frac{3}{4}x^{2}=5+7$$

Add 5 and 7 to get 12.

$$\frac{3}{4}x^{2}=12$$

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

$$x^{2}=12\times \frac{4}{3}$$

Multiply 12 and \frac{4}{3} to get 16.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

Steps Using the Quadratic Formula

Subtract 5 from both sides.

$$\frac{3}{4}x^{2}-7-5=0$$

Subtract 5 from -7 to get -12.

$$\frac{3}{4}x^{2}-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{4} for a, 0 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{3}{4}\left(-12\right)}}{2\times \frac{3}{4}}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times \frac{3}{4}\left(-12\right)}}{2\times \frac{3}{4}}$$

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

$$x=\frac{0±\sqrt{-3\left(-12\right)}}{2\times \frac{3}{4}}$$

Multiply -3 times -12.

$$x=\frac{0±\sqrt{36}}{2\times \frac{3}{4}}$$

Take the square root of 36.

$$x=\frac{0±6}{2\times \frac{3}{4}}$$

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

$$x=\frac{0±6}{\frac{3}{2}}$$

Now solve the equation x=\frac{0±6}{\frac{3}{2}} when ± is plus. Divide 6 by \frac{3}{2} by multiplying 6 by the reciprocal of \frac{3}{2}.

$$x=4$$

Now solve the equation x=\frac{0±6}{\frac{3}{2}} when ± is minus. Divide -6 by \frac{3}{2} by multiplying -6 by the reciprocal of \frac{3}{2}.

$$x=-4$$

The equation is now solved.

$$x=4 x=-4$$