-150+90t-4.90t^2=0

Solution for -150+90t-4.90t^2=0 equation:

-150+90t-4.90t^2=0

We add all the numbers together, and all the variables

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

a = -4.9; b = 90; c = -150;
Δ = b2-4ac
Δ = 902-4·(-4.9)·(-150)
Δ = 5160
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:

$\sqrt{\Delta}=\sqrt{5160}=\sqrt{4*1290}=\sqrt{4}*\sqrt{1290}=2\sqrt{1290}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-2\sqrt{1290}}{2*-4.9}=\frac{-90-2\sqrt{1290}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+2\sqrt{1290}}{2*-4.9}=\frac{-90+2\sqrt{1290}}{-9.8}
$