-8t^2+18t+5=0

Solution for -8t^2+18t+5=0 equation:

-8t^2+18t+5=0

a = -8; b = 18; c = +5;
Δ = b2-4ac
Δ = 182-4·(-8)·5
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*-8}=\frac{-40}{-16}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*-8}=\frac{4}{-16}
=-1/4
$