Solve 12x^2=36

$$12 { x }^{ 2 } =36$$

Solve for x

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

Steps by Finding Square Root

Divide both sides by 12.

$$x^{2}=\frac{36}{12}$$

Divide 36 by 12 to get 3.

$$x^{2}=3$$

Take the square root of both sides of the equation.

$$x=\sqrt{3} x=-\sqrt{3}$$

Steps Using the Quadratic Formula

Divide both sides by 12.

$$x^{2}=\frac{36}{12}$$

Divide 36 by 12 to get 3.

$$x^{2}=3$$

Subtract 3 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 0.

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

Multiply -4 times -3.

$$x=\frac{0±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{0±2\sqrt{3}}{2}$$

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

$$x=\sqrt{3}$$

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

$$x=-\sqrt{3}$$

The equation is now solved.

$$x=\sqrt{3} x=-\sqrt{3}$$