Solve Using the Quadratic Formula (t+4)^2-13(t+4)+42=0

${(t + 4)}^{2} - 13 (t + 4) + 42 = 0$

Step 1

Simplify the left side.

Simplify ${(t + 4)}^{2} - 13 (t + 4) + 42$ .

Simplify each term.

Apply the distributive property.

${(t + 4)}^{2} - 13 t - 13 \cdot 4 + 42 = 0$

Multiply $- 13$ by $4$ .

${(t + 4)}^{2} - 13 t - 52 + 42 = 0$

Add $- 52$ and $42$ .

${(t + 4)}^{2} - 13 t - 10 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = - 5$ , and $c = 6$ into the quadratic formula and solve for $t$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot 6)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$t = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 6$ .

Multiply $- 4$ by $1$ .

$t = \frac{5 \pm \sqrt{25 - 4 \cdot 6}}{2 \cdot 1}$

Multiply $- 4$ by $6$ .

$t = \frac{5 \pm \sqrt{25 - 24}}{2 \cdot 1}$

Subtract $24$ from $25$ .

$t = \frac{5 \pm \sqrt{1}}{2 \cdot 1}$

Any root of $1$ is $1$ .

$t = \frac{5 \pm 1}{2 \cdot 1}$

Multiply $2$ by $1$ .

$t = \frac{5 \pm 1}{2}$

Step 5

The final answer is the combination of both solutions.

$t = 3, 2$