Solve a^2+3^2=16

$$a ^ { 2 } + 3 ^ { 2 } = 16$$

Solve for a

$$a=\sqrt{7}\approx 2.645751311$$

$$a=-\sqrt{7}\approx -2.645751311$$

Steps by Finding Square Root

Calculate 3 to the power of 2 and get 9.

$$a^{2}+9=16$$

Subtract 9 from both sides.

$$a^{2}=16-9$$

Subtract 9 from 16 to get 7.

$$a^{2}=7$$

Take the square root of both sides of the equation.

$$a=\sqrt{7} a=-\sqrt{7}$$

Steps Using the Quadratic Formula

Calculate 3 to the power of 2 and get 9.

$$a^{2}+9=16$$

Subtract 16 from both sides.

$$a^{2}+9-16=0$$

Subtract 16 from 9 to get -7.

$$a^{2}-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\left(-7\right)}}{2}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-7\right)}}{2}$$

Multiply -4 times -7.

$$a=\frac{0±\sqrt{28}}{2}$$

Take the square root of 28.

$$a=\frac{0±2\sqrt{7}}{2}$$

Now solve the equation a=\frac{0±2\sqrt{7}}{2} when ± is plus.

$$a=\sqrt{7}$$

Now solve the equation a=\frac{0±2\sqrt{7}}{2} when ± is minus.

$$a=-\sqrt{7}$$

The equation is now solved.

$$a=\sqrt{7} a=-\sqrt{7}$$