Y=2s^2+s-3

Solution for Y=2s^2+s-3 equation:

=2Y^2+Y-3

We move all terms to the left:

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

We get rid of parentheses

-2Y^2-Y+3=0

We add all the numbers together, and all the variables

-2Y^2-1Y+3=0

a = -2; b = -1; c = +3;
Δ = b2-4ac
Δ = -12-4·(-2)·3
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-2}=\frac{-4}{-4}
=1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-2}=\frac{6}{-4}
=-1+1/2
$