Solve 40p^2-90=0

$$40 p ^ { 2 } - 90 = 0$$

Solve for p

$$p = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$p = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Difference of Squares

Divide both sides by 10.

$$4p^{2}-9=0$$

Consider 4p^{2}-9. Rewrite 4p^{2}-9 as \left(2p\right)^{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).

$$\left(2p-3\right)\left(2p+3\right)=0$$

To find equation solutions, solve 2p-3=0 and 2p+3=0.

$$p=\frac{3}{2} p=-\frac{3}{2}$$

Steps by Finding Square Root

Add 90 to both sides. Anything plus zero gives itself.

$$40p^{2}=90$$

Divide both sides by 40.

$$p^{2}=\frac{90}{40}$$

Reduce the fraction \frac{90}{40} to lowest terms by extracting and canceling out 10.

$$p^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$p=\frac{3}{2} p=-\frac{3}{2}$$

Steps Using the Quadratic Formula

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.

$$40p^{2}-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, 0 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{0±\sqrt{0^{2}-4\times 40\left(-90\right)}}{2\times 40}$$

Square 0.

$$p=\frac{0±\sqrt{-4\times 40\left(-90\right)}}{2\times 40}$$

Multiply -4 times 40.

$$p=\frac{0±\sqrt{-160\left(-90\right)}}{2\times 40}$$

Multiply -160 times -90.

$$p=\frac{0±\sqrt{14400}}{2\times 40}$$

Take the square root of 14400.

$$p=\frac{0±120}{2\times 40}$$

Multiply 2 times 40.

$$p=\frac{0±120}{80}$$

Now solve the equation p=\frac{0±120}{80} when ± is plus. Reduce the fraction \frac{120}{80} to lowest terms by extracting and canceling out 40.

$$p=\frac{3}{2}$$

Now solve the equation p=\frac{0±120}{80} when ± is minus. Reduce the fraction \frac{-120}{80} to lowest terms by extracting and canceling out 40.

$$p=-\frac{3}{2}$$

The equation is now solved.

$$p=\frac{3}{2} p=-\frac{3}{2}$$