1200+150t-15t^2=0

Solution for 1200+150t-15t^2=0 equation:

1200+150t-15t^2=0

a = -15; b = 150; c = +1200;
Δ = b2-4ac
Δ = 1502-4·(-15)·1200
Δ = 94500
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:

$\sqrt{\Delta}=\sqrt{94500}=\sqrt{900*105}=\sqrt{900}*\sqrt{105}=30\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-30\sqrt{105}}{2*-15}=\frac{-150-30\sqrt{105}}{-30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+30\sqrt{105}}{2*-15}=\frac{-150+30\sqrt{105}}{-30}
$