0.9t^2+4.5t-1500=0

Solution for 0.9t^2+4.5t-1500=0 equation:

0.9t^2+4.5t-1500=0

a = 0.9; b = 4.5; c = -1500;
Δ = b2-4ac
Δ = 4.52-4·0.9·(-1500)
Δ = 5420.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4.5)-\sqrt{5420.25}}{2*0.9}=\frac{-4.5-\sqrt{5420.25}}{1.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.5)+\sqrt{5420.25}}{2*0.9}=\frac{-4.5+\sqrt{5420.25}}{1.8}
$