Solve 36=frac{3}2^2+x^2

$$36 = \frac { 3 } { 2 } ^ { 2 } + x ^ { 2 }$$

Solve for x

$$x = \frac{3 \sqrt{15}}{2} \approx 5.809475019$$

$$x = -\frac{3 \sqrt{15}}{2} \approx -5.809475019$$

Steps by Finding Square Root

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$36=\frac{9}{4}+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{9}{4}+x^{2}=36$$

Subtract \frac{9}{4} from both sides.

$$x^{2}=36-\frac{9}{4}$$

Subtract \frac{9}{4} from 36 to get \frac{135}{4}.

$$x^{2}=\frac{135}{4}$$

Take the square root of both sides of the equation.

$$x=\frac{3\sqrt{15}}{2} x=-\frac{3\sqrt{15}}{2}$$

Steps Using the Quadratic Formula

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$36=\frac{9}{4}+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{9}{4}+x^{2}=36$$

Subtract 36 from both sides.

$$\frac{9}{4}+x^{2}-36=0$$

Subtract 36 from \frac{9}{4} to get -\frac{135}{4}.

$$-\frac{135}{4}+x^{2}=0$$

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.

$$x^{2}-\frac{135}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{135}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{135}{4}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{135}{4}\right)}}{2}$$

Multiply -4 times -\frac{135}{4}.

$$x=\frac{0±\sqrt{135}}{2}$$

Take the square root of 135.

$$x=\frac{0±3\sqrt{15}}{2}$$

Now solve the equation x=\frac{0±3\sqrt{15}}{2} when ± is plus.

$$x=\frac{3\sqrt{15}}{2}$$

Now solve the equation x=\frac{0±3\sqrt{15}}{2} when ± is minus.

$$x=-\frac{3\sqrt{15}}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{15}}{2} x=-\frac{3\sqrt{15}}{2}$$