16t^2-220t+624=0

Solution for 16t^2-220t+624=0 equation:

16t^2-220t+624=0

a = 16; b = -220; c = +624;
Δ = b2-4ac
Δ = -2202-4·16·624
Δ = 8464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8464}=92$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-220)-92}{2*16}=\frac{128}{32}
=4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-220)+92}{2*16}=\frac{312}{32}
=9+3/4
$