-16t^2+150t+100=0

Solution for -16t^2+150t+100=0 equation:

-16t^2+150t+100=0

a = -16; b = 150; c = +100;
Δ = b2-4ac
Δ = 1502-4·(-16)·100
Δ = 28900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{28900}=170$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-170}{2*-16}=\frac{-320}{-32}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+170}{2*-16}=\frac{20}{-32}
=-5/8
$