-2a^2+2a+8=0

Solution for -2a^2+2a+8=0 equation:

-2a^2+2a+8=0

a = -2; b = 2; c = +8;
Δ = b2-4ac
Δ = 22-4·(-2)·8
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{17}}{2*-2}=\frac{-2-2\sqrt{17}}{-4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{17}}{2*-2}=\frac{-2+2\sqrt{17}}{-4}
$