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

Solution for (t)=2t^2+3t-2 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = -2; c = +2;
Δ = b2-4ac
Δ = -22-4·(-2)·2
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{5}}{2*-2}=\frac{2-2\sqrt{5}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{5}}{2*-2}=\frac{2+2\sqrt{5}}{-4}
$