-4.9*t^2+90*t+150=0

Solution for -4.9*t^2+90*t+150=0 equation:

-4.9t^2+90t+150=0

a = -4.9; b = 90; c = +150;
Δ = b2-4ac
Δ = 902-4·(-4.9)·150
Δ = 11040
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:

$\sqrt{\Delta}=\sqrt{11040}=\sqrt{16*690}=\sqrt{16}*\sqrt{690}=4\sqrt{690}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-4\sqrt{690}}{2*-4.9}=\frac{-90-4\sqrt{690}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+4\sqrt{690}}{2*-4.9}=\frac{-90+4\sqrt{690}}{-9.8}
$