0=2t^2-2t-5

Solution for 0=2t^2-2t-5 equation:

0=2t^2-2t-5

We move all terms to the left:

0-(2t^2-2t-5)=0

We add all the numbers together, and all the variables

-(2t^2-2t-5)=0

We get rid of parentheses

-2t^2+2t+5=0

a = -2; b = 2; c = +5;
Δ = b2-4ac
Δ = 22-4·(-2)·5
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{11}}{2*-2}=\frac{-2-2\sqrt{11}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{11}}{2*-2}=\frac{-2+2\sqrt{11}}{-4}
$