Y=10x^2+8x-3

Solution for Y=10x^2+8x-3 equation:

=10Y^2+8Y-3

We move all terms to the left:

-(10Y^2+8Y-3)=0

We get rid of parentheses

-10Y^2-8Y+3=0

a = -10; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·(-10)·3
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{46}}{2*-10}=\frac{8-2\sqrt{46}}{-20}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{46}}{2*-10}=\frac{8+2\sqrt{46}}{-20}
$