0=-2t^2+4t+88

Solution for 0=-2t^2+4t+88 equation:

0=-2t^2+4t+88

We move all terms to the left:

0-(-2t^2+4t+88)=0

We add all the numbers together, and all the variables

-(-2t^2+4t+88)=0

We get rid of parentheses

2t^2-4t-88=0

a = 2; b = -4; c = -88;
Δ = b2-4ac
Δ = -42-4·2·(-88)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12\sqrt{5}}{2*2}=\frac{4-12\sqrt{5}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12\sqrt{5}}{2*2}=\frac{4+12\sqrt{5}}{4}
$