x2+-40x+-48=0

Solution for x2+-40x+-48=0 equation:

x2+-40x+-48=0

We add all the numbers together, and all the variables

x^2-40x=0

a = 1; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·1·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*1}=\frac{80}{2}
=40
$