-2x^2-20x+5=0

Solution for -2x^2-20x+5=0 equation:

-2x^2-20x+5=0

a = -2; b = -20; c = +5;
Δ = b2-4ac
Δ = -202-4·(-2)·5
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{110}}{2*-2}=\frac{20-2\sqrt{110}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{110}}{2*-2}=\frac{20+2\sqrt{110}}{-4}
$