Solve Using the Quadratic Formula 4y^2-25=0

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

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 4$ , $b = 0$ , and $c = - 25$ into the quadratic formula and solve for $y$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (4 \cdot - 25)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$y = \frac{0 \pm \sqrt{0 - 4 \cdot 4 \cdot - 25}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 25$ .

Multiply $- 4$ by $4$ .

$y = \frac{0 \pm \sqrt{0 - 16 \cdot - 25}}{2 \cdot 4}$

Multiply $- 16$ by $- 25$ .

$y = \frac{0 \pm \sqrt{0 + 400}}{2 \cdot 4}$

Add $0$ and $400$ .

$y = \frac{0 \pm \sqrt{400}}{2 \cdot 4}$

Rewrite $400$ as $20^{2}$ .

$y = \frac{0 \pm \sqrt{20^{2}}}{2 \cdot 4}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \frac{0 \pm 20}{2 \cdot 4}$

Multiply $2$ by $4$ .

$y = \frac{0 \pm 20}{8}$

Simplify $\frac{0 \pm 20}{8}$ .

$y = \frac{\pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

$y = \frac{5}{2}, - \frac{5}{2}$