-4x^2-9600+400x=0

Solution for -4x^2-9600+400x=0 equation:

-4x^2-9600+400x=0

a = -4; b = 400; c = -9600;
Δ = b2-4ac
Δ = 4002-4·(-4)·(-9600)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(400)-80}{2*-4}=\frac{-480}{-8}
=+60
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(400)+80}{2*-4}=\frac{-320}{-8}
=+40
$