Y=10x+2x^2

Solution for Y=10x+2x^2 equation:

=10Y+2Y^2

We move all terms to the left:

-(10Y+2Y^2)=0

We get rid of parentheses

-2Y^2-10Y=0

a = -2; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·(-2)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*-2}=\frac{0}{-4}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*-2}=\frac{20}{-4}
=-5
$