0.9t^2+3t-1700=0

Solution for 0.9t^2+3t-1700=0 equation:

0.9t^2+3t-1700=0

a = 0.9; b = 3; c = -1700;
Δ = b2-4ac
Δ = 32-4·0.9·(-1700)
Δ = 6129
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:

$\sqrt{\Delta}=\sqrt{6129}=\sqrt{9*681}=\sqrt{9}*\sqrt{681}=3\sqrt{681}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{681}}{2*0.9}=\frac{-3-3\sqrt{681}}{1.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{681}}{2*0.9}=\frac{-3+3\sqrt{681}}{1.8}
$