Y=-5+3x+2x^2

Solution for Y=-5+3x+2x^2 equation:

=-5+3Y+2Y^2

We move all terms to the left:

-(-5+3Y+2Y^2)=0

We get rid of parentheses

-2Y^2-3Y+5=0

a = -2; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-2)·5
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*-2}=\frac{-4}{-4}
=1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-2}=\frac{10}{-4}
=-2+1/2
$