-4t^2+48t-80=0

Solution for -4t^2+48t-80=0 equation:

-4t^2+48t-80=0

a = -4; b = 48; c = -80;
Δ = b2-4ac
Δ = 482-4·(-4)·(-80)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-32}{2*-4}=\frac{-80}{-8}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+32}{2*-4}=\frac{-16}{-8}
=+2
$