Solve P^2-5=4

$$P ^ { 2 } - 5 = 4$$

Solve for P

$$P=3$$

$$P=-3$$

Steps Using Difference of Squares

Subtract 4 from both sides.

$$P^{2}-5-4=0$$

Subtract 4 from -5 to get -9.

$$P^{2}-9=0$$

Consider P^{2}-9. Rewrite P^{2}-9 as P^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(P-3\right)\left(P+3\right)=0$$

To find equation solutions, solve P-3=0 and P+3=0.

$$P=3 P=-3$$

Steps by Finding Square Root

Add 5 to both sides.

$$P^{2}=4+5$$

Add 4 and 5 to get 9.

$$P^{2}=9$$

Take the square root of both sides of the equation.

$$P=3 P=-3$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$P^{2}-5-4=0$$

Subtract 4 from -5 to get -9.

$$P^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$P=\frac{0±\sqrt{0^{2}-4\left(-9\right)}}{2}$$

Square 0.

$$P=\frac{0±\sqrt{-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$P=\frac{0±\sqrt{36}}{2}$$

Take the square root of 36.

$$P=\frac{0±6}{2}$$

Now solve the equation P=\frac{0±6}{2} when ± is plus. Divide 6 by 2.

$$P=3$$

Now solve the equation P=\frac{0±6}{2} when ± is minus. Divide -6 by 2.

$$P=-3$$

The equation is now solved.

$$P=3 P=-3$$