0=-4.905x^2+90x+150

Solution for 0=-4.905x^2+90x+150 equation:

0=-4.905x^2+90x+150

We move all terms to the left:

0-(-4.905x^2+90x+150)=0

We add all the numbers together, and all the variables

-(-4.905x^2+90x+150)=0

We get rid of parentheses

4.905x^2-90x-150=0

a = 4.905; b = -90; c = -150;
Δ = b2-4ac
Δ = -902-4·4.905·(-150)
Δ = 11043
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{11043}=\sqrt{9*1227}=\sqrt{9}*\sqrt{1227}=3\sqrt{1227}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-3\sqrt{1227}}{2*4.905}=\frac{90-3\sqrt{1227}}{9.81}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+3\sqrt{1227}}{2*4.905}=\frac{90+3\sqrt{1227}}{9.81}
$