Solve for r 2r^2+7r-1=-3r^2

$2 r^{2} + 7 r - 1 = - 3 r^{2}$

Move all terms containing $r$ to the left side of the equation.

Add $3 r^{2}$ to both sides of the equation.

$2 r^{2} + 7 r - 1 + 3 r^{2} = 0$

Add $2 r^{2}$ and $3 r^{2}$ .

$5 r^{2} + 7 r - 1 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 5$ , $b = 7$ , and $c = - 1$ into the quadratic formula and solve for $r$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (5 \cdot - 1)}}{2 \cdot 5}$

Simplify.

Simplify the numerator.

Raise $7$ to the power of $2$ .

$r = \frac{- 7 \pm \sqrt{49 - 4 \cdot 5 \cdot - 1}}{2 \cdot 5}$

Multiply $- 4$ by $5$ .

$r = \frac{- 7 \pm \sqrt{49 - 20 \cdot - 1}}{2 \cdot 5}$

Multiply $- 20$ by $- 1$ .

$r = \frac{- 7 \pm \sqrt{49 + 20}}{2 \cdot 5}$

Add $49$ and $20$ .

$r = \frac{- 7 \pm \sqrt{69}}{2 \cdot 5}$

Multiply $2$ by $5$ .

$r = \frac{- 7 \pm \sqrt{69}}{10}$

The final answer is the combination of both solutions.

$r = - \frac{7 - \sqrt{69}}{10}, - \frac{7 + \sqrt{69}}{10}$

The result can be shown in multiple forms.

Exact Form:

$r = - \frac{7 - \sqrt{69}}{10}, - \frac{7 + \sqrt{69}}{10}$

Decimal Form:

$r = 0.13066238 \dots, - 1.53066238 \dots$