8t^2-22t-56=0

Solution for 8t^2-22t-56=0 equation:

8t^2-22t-56=0

a = 8; b = -22; c = -56;
Δ = b2-4ac
Δ = -222-4·8·(-56)
Δ = 2276
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:

$\sqrt{\Delta}=\sqrt{2276}=\sqrt{4*569}=\sqrt{4}*\sqrt{569}=2\sqrt{569}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{569}}{2*8}=\frac{22-2\sqrt{569}}{16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{569}}{2*8}=\frac{22+2\sqrt{569}}{16}
$