Solve for r 1-5r=-2r^2

$1 - 5 r = - 2 r^{2}$

Add $2 r^{2}$ to both sides of the equation.

$1 - 5 r + 2 r^{2} = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 5$ , and $c = 1$ into the quadratic formula and solve for $r$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $2$ .

$r = \frac{5 \pm \sqrt{25 - 8 \cdot 1}}{2 \cdot 2}$

Multiply $- 8$ by $1$ .

$r = \frac{5 \pm \sqrt{25 - 8}}{2 \cdot 2}$

Subtract $8$ from $25$ .

$r = \frac{5 \pm \sqrt{17}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$r = \frac{5 \pm \sqrt{17}}{4}$

The final answer is the combination of both solutions.

$r = \frac{5 + \sqrt{17}}{4}, \frac{5 - \sqrt{17}}{4}$

The result can be shown in multiple forms.

Exact Form:

$r = \frac{5 + \sqrt{17}}{4}, \frac{5 - \sqrt{17}}{4}$

Decimal Form:

$r = 2.28077640 \dots, 0.21922359 \dots$