0=150+90t+4.9t^2

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

0=150+90t+4.9t^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-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}
$