Find the Roots (Zeros) -2r^2+r+7=0

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = - 2$ , $b = 1$ , and $c = 7$ into the quadratic formula and solve for $r$ .

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

Simplify.

Simplify the numerator.

One to any power is one.

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

Multiply $- 4$ by $- 2$ .

$r = \frac{- 1 \pm \sqrt{1 + 8 \cdot 7}}{2 \cdot - 2}$

Multiply $8$ by $7$ .

$r = \frac{- 1 \pm \sqrt{1 + 56}}{2 \cdot - 2}$

Add $1$ and $56$ .

$r = \frac{- 1 \pm \sqrt{57}}{2 \cdot - 2}$

Multiply $2$ by $- 2$ .

$r = \frac{- 1 \pm \sqrt{57}}{- 4}$

Simplify $\frac{- 1 \pm \sqrt{57}}{- 4}$ .

$r = \frac{1 \pm \sqrt{57}}{4}$

The result can be shown in multiple forms.

Exact Form:

$r = \frac{1 \pm \sqrt{57}}{4}$

Decimal Form:

$r = 2.13745860 \dots, - 1.63745860 \dots$