Solve a^2=99

$$a ^ { 2 } = 99$$

Solve for a

$$a=3\sqrt{11}\approx 9.949874371$$

$$a=-3\sqrt{11}\approx -9.949874371$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$a=3\sqrt{11} a=-3\sqrt{11}$$

Steps Using the Quadratic Formula

Subtract 99 from both sides.

$$a^{2}-99=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -99 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\left(-99\right)}}{2}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-99\right)}}{2}$$

Multiply -4 times -99.

$$a=\frac{0±\sqrt{396}}{2}$$

Take the square root of 396.

$$a=\frac{0±6\sqrt{11}}{2}$$

Now solve the equation a=\frac{0±6\sqrt{11}}{2} when ± is plus.

$$a=3\sqrt{11}$$

Now solve the equation a=\frac{0±6\sqrt{11}}{2} when ± is minus.

$$a=-3\sqrt{11}$$

The equation is now solved.

$$a=3\sqrt{11} a=-3\sqrt{11}$$