Solve 4c^2=25

$$4 c ^ { 2 } = 25$$

Solve for c

$$c = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$c = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps Using Difference of Squares

Divide both sides by 4.

$$c^{2}=\frac{25}{4}$$

Subtract \frac{25}{4} from both sides.

$$c^{2}-\frac{25}{4}=0$$

Multiply both sides by 4.

$$4c^{2}-25=0$$

Consider 4c^{2}-25. Rewrite 4c^{2}-25 as \left(2c\right)^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2c-5\right)\left(2c+5\right)=0$$

To find equation solutions, solve 2c-5=0 and 2c+5=0.

$$c=\frac{5}{2} c=-\frac{5}{2}$$

Steps by Finding Square Root

Divide both sides by 4.

$$c^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$c=\frac{5}{2} c=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Divide both sides by 4.

$$c^{2}=\frac{25}{4}$$

Subtract \frac{25}{4} from both sides.

$$c^{2}-\frac{25}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{25}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{0±\sqrt{0^{2}-4\left(-\frac{25}{4}\right)}}{2}$$

Square 0.

$$c=\frac{0±\sqrt{-4\left(-\frac{25}{4}\right)}}{2}$$

Multiply -4 times -\frac{25}{4}.

$$c=\frac{0±\sqrt{25}}{2}$$

Take the square root of 25.

$$c=\frac{0±5}{2}$$

Now solve the equation c=\frac{0±5}{2} when ± is plus. Divide 5 by 2.

$$c=\frac{5}{2}$$

Now solve the equation c=\frac{0±5}{2} when ± is minus. Divide -5 by 2.

$$c=-\frac{5}{2}$$

The equation is now solved.

$$c=\frac{5}{2} c=-\frac{5}{2}$$