Solve Using the Quadratic Formula t^2-25=0

$t^{2} - 25 = 0$

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 $a = 1$ , $b = 0$ , and $c = - 25$ into the quadratic formula and solve for $t$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (1 \cdot - 25)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$t = \frac{0 \pm \sqrt{0 - 4 \cdot 1 \cdot - 25}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 25$ .

Multiply $- 4$ by $1$ .

$t = \frac{0 \pm \sqrt{0 - 4 \cdot - 25}}{2 \cdot 1}$

Multiply $- 4$ by $- 25$ .

$t = \frac{0 \pm \sqrt{0 + 100}}{2 \cdot 1}$

Add $0$ and $100$ .

$t = \frac{0 \pm \sqrt{100}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$t = \frac{0 \pm \sqrt{10^{2}}}{2 \cdot 1}$

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

$t = \frac{0 \pm 10}{2 \cdot 1}$

Multiply $2$ by $1$ .

$t = \frac{0 \pm 10}{2}$

Simplify $\frac{0 \pm 10}{2}$ .

$t = \pm 5$

Step 4

The final answer is the combination of both solutions.

$t = 5, - 5$