-2x2+20x-46=0

Solution for -2x2+20x-46=0 equation:

-2x^2+20x-46=0

a = -2; b = 20; c = -46;
Δ = b2-4ac
Δ = 202-4·(-2)·(-46)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{2}}{2*-2}=\frac{-20-4\sqrt{2}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{2}}{2*-2}=\frac{-20+4\sqrt{2}}{-4}
$