Y=2x^2+5x-3x^2+4x+3

Solution for Y=2x^2+5x-3x^2+4x+3 equation:

=2Y^2+5Y-3Y^2+4Y+3

We move all terms to the left:

-(2Y^2+5Y-3Y^2+4Y+3)=0

We get rid of parentheses

-2Y^2+3Y^2-5Y-4Y-3=0

We add all the numbers together, and all the variables

Y^2-9Y-3=0

a = 1; b = -9; c = -3;
Δ = b2-4ac
Δ = -92-4·1·(-3)
Δ = 93
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{-(-9)-\sqrt{93}}{2*1}=\frac{9-\sqrt{93}}{2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{93}}{2*1}=\frac{9+\sqrt{93}}{2}
$