-2x2+48x-160=0

Solution for -2x2+48x-160=0 equation:

-2x^2+48x-160=0

a = -2; b = 48; c = -160;
Δ = b2-4ac
Δ = 482-4·(-2)·(-160)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-32}{2*-2}=\frac{-80}{-4}
=+20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+32}{2*-2}=\frac{-16}{-4}
=+4
$