Solve Using the Quadratic Formula t^2+5t+5=0

$t^{2} + 5 t + 5 = 0$

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

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

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot 5$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $5$ .

$t = \frac{- 5 \pm \sqrt{25 - 20}}{2 \cdot 1}$

Subtract $20$ from $25$ .

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

Multiply $2$ by $1$ .

$t = \frac{- 5 \pm \sqrt{5}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$t = \frac{- 5 \pm \sqrt{5}}{2}$

Decimal Form:

$t = - 1.38196601 \dots, - 3.61803398 \dots$