2+40t-17t^2=0

Solution for 2+40t-17t^2=0 equation:

2+40t-17t^2=0

a = -17; b = 40; c = +2;
Δ = b2-4ac
Δ = 402-4·(-17)·2
Δ = 1736
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:

$\sqrt{\Delta}=\sqrt{1736}=\sqrt{4*434}=\sqrt{4}*\sqrt{434}=2\sqrt{434}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{434}}{2*-17}=\frac{-40-2\sqrt{434}}{-34}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{434}}{2*-17}=\frac{-40+2\sqrt{434}}{-34}
$