-2x^2+2x+33=0

Solution for -2x^2+2x+33=0 equation:

-2x^2+2x+33=0

a = -2; b = 2; c = +33;
Δ = b2-4ac
Δ = 22-4·(-2)·33
Δ = 268
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{67}}{2*-2}=\frac{-2-2\sqrt{67}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{67}}{2*-2}=\frac{-2+2\sqrt{67}}{-4}
$