Solve Using the Quadratic Formula r^2+3r=1

$r^{2} + 3 r = 1$

Step 1

Subtract $1$ from both sides of the equation.

$r^{2} + 3 r - 1 = 0$

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

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Add $9$ and $4$ .

$r = \frac{- 3 \pm \sqrt{13}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{- 3 \pm \sqrt{13}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$r = \frac{- 3 \pm \sqrt{13}}{2}$

Decimal Form:

$r = 0.30277563 \dots, - 3.30277563 \dots$