Solve Using the Quadratic Formula r^2+3r-5=0

$r^{2} + 3 r - 5 = 0$

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 $a = 1$ , $b = 3$ , and $c = - 5$ into the quadratic formula and solve for $r$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 5)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $- 5$ .

$r = \frac{- 3 \pm \sqrt{9 + 20}}{2 \cdot 1}$

Add $9$ and $20$ .

$r = \frac{- 3 \pm \sqrt{29}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{- 3 \pm \sqrt{29}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$r = \frac{- 3 \pm \sqrt{29}}{2}$

Decimal Form:

$r = 1.19258240 \dots, - 4.19258240 \dots$