Solve 9+x^2=36

$$9 + x ^ { 2 } = 36$$

Solve for x

$$x=3\sqrt{3}\approx 5.196152423$$

$$x=-3\sqrt{3}\approx -5.196152423$$

Steps by Finding Square Root

Subtract 9 from both sides.

$$x^{2}=36-9$$

Subtract 9 from 36 to get 27.

$$x^{2}=27$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Subtract 36 from both sides.

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

Subtract 36 from 9 to get -27.

$$-27+x^{2}=0$$

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.

$$x^{2}-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-27\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-27\right)}}{2}$$

Multiply -4 times -27.

$$x=\frac{0±\sqrt{108}}{2}$$

Take the square root of 108.

$$x=\frac{0±6\sqrt{3}}{2}$$

Now solve the equation x=\frac{0±6\sqrt{3}}{2} when ± is plus.

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

Now solve the equation x=\frac{0±6\sqrt{3}}{2} when ± is minus.

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

The equation is now solved.

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