Solve 4+a^2=16

$$4 + a ^ { 2 } = 16$$

Solve for a

$$a=2\sqrt{3}\approx 3.464101615$$

$$a=-2\sqrt{3}\approx -3.464101615$$

Steps by Finding Square Root

Subtract 4 from both sides.

$$a^{2}=16-4$$

Subtract 4 from 16 to get 12.

$$a^{2}=12$$

Take the square root of both sides of the equation.

$$a=2\sqrt{3} a=-2\sqrt{3}$$

Steps Using the Quadratic Formula

Subtract 16 from both sides.

$$4+a^{2}-16=0$$

Subtract 16 from 4 to get -12.

$$-12+a^{2}=0$$

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.

$$a^{2}-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\left(-12\right)}}{2}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$a=\frac{0±\sqrt{48}}{2}$$

Take the square root of 48.

$$a=\frac{0±4\sqrt{3}}{2}$$

Now solve the equation a=\frac{0±4\sqrt{3}}{2} when ± is plus.

$$a=2\sqrt{3}$$

Now solve the equation a=\frac{0±4\sqrt{3}}{2} when ± is minus.

$$a=-2\sqrt{3}$$

The equation is now solved.

$$a=2\sqrt{3} a=-2\sqrt{3}$$