11t^2+17t-240=0

Solution for 11t^2+17t-240=0 equation:

11t^2+17t-240=0

a = 11; b = 17; c = -240;
Δ = b2-4ac
Δ = 172-4·11·(-240)
Δ = 10849
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{-(17)-\sqrt{10849}}{2*11}=\frac{-17-\sqrt{10849}}{22}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{10849}}{2*11}=\frac{-17+\sqrt{10849}}{22}
$