8t^2+4t-68=0

Solution for 8t^2+4t-68=0 equation:

8t^2+4t-68=0

a = 8; b = 4; c = -68;
Δ = b2-4ac
Δ = 42-4·8·(-68)
Δ = 2192
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:

$\sqrt{\Delta}=\sqrt{2192}=\sqrt{16*137}=\sqrt{16}*\sqrt{137}=4\sqrt{137}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{137}}{2*8}=\frac{-4-4\sqrt{137}}{16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{137}}{2*8}=\frac{-4+4\sqrt{137}}{16}
$