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

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

-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}
$