-16x^2+160x-300=0

Solution for -16x^2+160x-300=0 equation:

-16x^2+160x-300=0

a = -16; b = 160; c = -300;
Δ = b2-4ac
Δ = 1602-4·(-16)·(-300)
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{-(160)-80}{2*-16}=\frac{-240}{-32}
=7+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+80}{2*-16}=\frac{-80}{-32}
=2+1/2
$