-16x^2+144x-320=0

Solution for -16x^2+144x-320=0 equation:

-16x^2+144x-320=0

a = -16; b = 144; c = -320;
Δ = b2-4ac
Δ = 1442-4·(-16)·(-320)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(144)-16}{2*-16}=\frac{-160}{-32}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(144)+16}{2*-16}=\frac{-128}{-32}
=+4
$