0=2t^2+3t-20

Solution for 0=2t^2+3t-20 equation:

0=2t^2+3t-20

We move all terms to the left:

0-(2t^2+3t-20)=0

We add all the numbers together, and all the variables

-(2t^2+3t-20)=0

We get rid of parentheses

-2t^2-3t+20=0

a = -2; b = -3; c = +20;
Δ = b2-4ac
Δ = -32-4·(-2)·20
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*-2}=\frac{-10}{-4}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*-2}=\frac{16}{-4}
=-4
$