5t^2-8t-11=0

Solution for 5t^2-8t-11=0 equation:

5t^2-8t-11=0

a = 5; b = -8; c = -11;
Δ = b2-4ac
Δ = -82-4·5·(-11)
Δ = 284
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:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{71}}{2*5}=\frac{8-2\sqrt{71}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{71}}{2*5}=\frac{8+2\sqrt{71}}{10}
$