0=8t^2-22t-20

Solution for 0=8t^2-22t-20 equation:

0=8t^2-22t-20

We move all terms to the left:

0-(8t^2-22t-20)=0

We add all the numbers together, and all the variables

-(8t^2-22t-20)=0

We get rid of parentheses

-8t^2+22t+20=0

a = -8; b = 22; c = +20;
Δ = b2-4ac
Δ = 222-4·(-8)·20
Δ = 1124
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:

$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{281}}{2*-8}=\frac{-22-2\sqrt{281}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{281}}{2*-8}=\frac{-22+2\sqrt{281}}{-16}
$