Solve p^2+4=11

$$p ^ { 2 } + 4 = 11$$

Solve for p

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

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

Steps by Finding Square Root

Subtract 4 from both sides.

$$p^{2}=11-4$$

Subtract 4 from 11 to get 7.

$$p^{2}=7$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Subtract 11 from both sides.

$$p^{2}+4-11=0$$

Subtract 11 from 4 to get -7.

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

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}.

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

Square 0.

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

Multiply -4 times -7.

$$p=\frac{0±\sqrt{28}}{2}$$

Take the square root of 28.

$$p=\frac{0±2\sqrt{7}}{2}$$

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

$$p=\sqrt{7}$$

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

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

The equation is now solved.

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