40t^2+80t-90=0

Solution for 40t^2+80t-90=0 equation:

40t^2+80t-90=0

a = 40; b = 80; c = -90;
Δ = b2-4ac
Δ = 802-4·40·(-90)
Δ = 20800
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:

$\sqrt{\Delta}=\sqrt{20800}=\sqrt{1600*13}=\sqrt{1600}*\sqrt{13}=40\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-40\sqrt{13}}{2*40}=\frac{-80-40\sqrt{13}}{80}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+40\sqrt{13}}{2*40}=\frac{-80+40\sqrt{13}}{80}
$