Y=2x^2+3x-15

Solution for Y=2x^2+3x-15 equation:

=2Y^2+3Y-15

We move all terms to the left:

-(2Y^2+3Y-15)=0

We get rid of parentheses

-2Y^2-3Y+15=0

a = -2; b = -3; c = +15;
Δ = b2-4ac
Δ = -32-4·(-2)·15
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{129}}{2*-2}=\frac{3-\sqrt{129}}{-4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*-2}=\frac{3+\sqrt{129}}{-4}
$