16t^2-95t+120=0

Solution for 16t^2-95t+120=0 equation:

16t^2-95t+120=0

a = 16; b = -95; c = +120;
Δ = b2-4ac
Δ = -952-4·16·120
Δ = 1345
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-\sqrt{1345}}{2*16}=\frac{95-\sqrt{1345}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+\sqrt{1345}}{2*16}=\frac{95+\sqrt{1345}}{32}
$