0=-2t^2-2t+24

Solution for 0=-2t^2-2t+24 equation:

0=-2t^2-2t+24

We move all terms to the left:

0-(-2t^2-2t+24)=0

We add all the numbers together, and all the variables

-(-2t^2-2t+24)=0

We get rid of parentheses

2t^2+2t-24=0

a = 2; b = 2; c = -24;
Δ = b2-4ac
Δ = 22-4·2·(-24)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*2}=\frac{-16}{4}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*2}=\frac{12}{4}
=3
$