Solve Using the Quadratic Formula t^2-19t+18=0

$t^{2} - 19 t + 18 = 0$

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 $a = 1$ , $b = - 19$ , and $c = 18$ into the quadratic formula and solve for $t$ .

$\frac{19 \pm \sqrt{{(- 19)}^{2} - 4 \cdot (1 \cdot 18)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 19$ to the power of $2$ .

$t = \frac{19 \pm \sqrt{361 - 4 \cdot 1 \cdot 18}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 18$ .

Multiply $- 4$ by $1$ .

$t = \frac{19 \pm \sqrt{361 - 4 \cdot 18}}{2 \cdot 1}$

Multiply $- 4$ by $18$ .

$t = \frac{19 \pm \sqrt{361 - 72}}{2 \cdot 1}$

Subtract $72$ from $361$ .

$t = \frac{19 \pm \sqrt{289}}{2 \cdot 1}$

Rewrite $289$ as $17^{2}$ .

$t = \frac{19 \pm \sqrt{17^{2}}}{2 \cdot 1}$

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

$t = \frac{19 \pm 17}{2 \cdot 1}$

Multiply $2$ by $1$ .

$t = \frac{19 \pm 17}{2}$

Step 4

The final answer is the combination of both solutions.

$t = 18, 1$