16t^2-90t+120=0

Solution for 16t^2-90t+120=0 equation:

16t^2-90t+120=0

a = 16; b = -90; c = +120;
Δ = b2-4ac
Δ = -902-4·16·120
Δ = 420
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-2\sqrt{105}}{2*16}=\frac{90-2\sqrt{105}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+2\sqrt{105}}{2*16}=\frac{90+2\sqrt{105}}{32}
$