Y=10x^2+2x-0.5

Solution for Y=10x^2+2x-0.5 equation:

=10Y^2+2Y-0.5

We move all terms to the left:

-(10Y^2+2Y-0.5)=0

We get rid of parentheses

-10Y^2-2Y+0.5=0

a = -10; b = -2; c = +0.5;
Δ = b2-4ac
Δ = -22-4·(-10)·0.5
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{6}}{2*-10}=\frac{2-2\sqrt{6}}{-20}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{6}}{2*-10}=\frac{2+2\sqrt{6}}{-20}
$