8t^2+2t-15=0

Solution for 8t^2+2t-15=0 equation:

8t^2+2t-15=0

a = 8; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·8·(-15)
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{-(2)-22}{2*8}=\frac{-24}{16}
=-1+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*8}=\frac{20}{16}
=1+1/4
$