t^2+900t-16200=0

Solution for t^2+900t-16200=0 equation:

t^2+900t-16200=0

a = 1; b = 900; c = -16200;
Δ = b2-4ac
Δ = 9002-4·1·(-16200)
Δ = 874800
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:

$\sqrt{\Delta}=\sqrt{874800}=\sqrt{291600*3}=\sqrt{291600}*\sqrt{3}=540\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(900)-540\sqrt{3}}{2*1}=\frac{-900-540\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(900)+540\sqrt{3}}{2*1}=\frac{-900+540\sqrt{3}}{2}
$