Solve p^2-6=94

$$p ^ { 2 } - 6 = 94$$

Solve for p

$$p=10$$

$$p=-10$$

Steps Using Difference of Squares

Subtract 94 from both sides.

$$p^{2}-6-94=0$$

Subtract 94 from -6 to get -100.

$$p^{2}-100=0$$

Consider p^{2}-100. Rewrite p^{2}-100 as p^{2}-10^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(p-10\right)\left(p+10\right)=0$$

To find equation solutions, solve p-10=0 and p+10=0.

$$p=10 p=-10$$

Steps by Finding Square Root

Add 6 to both sides.

$$p^{2}=94+6$$

Add 94 and 6 to get 100.

$$p^{2}=100$$

Take the square root of both sides of the equation.

$$p=10 p=-10$$

Steps Using the Quadratic Formula

Subtract 94 from both sides.

$$p^{2}-6-94=0$$

Subtract 94 from -6 to get -100.

$$p^{2}-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{0±\sqrt{0^{2}-4\left(-100\right)}}{2}$$

Square 0.

$$p=\frac{0±\sqrt{-4\left(-100\right)}}{2}$$

Multiply -4 times -100.

$$p=\frac{0±\sqrt{400}}{2}$$

Take the square root of 400.

$$p=\frac{0±20}{2}$$

Now solve the equation p=\frac{0±20}{2} when ± is plus. Divide 20 by 2.

$$p=10$$

Now solve the equation p=\frac{0±20}{2} when ± is minus. Divide -20 by 2.

$$p=-10$$

The equation is now solved.

$$p=10 p=-10$$