Y=10x2+28x-3

Solution for Y=10x2+28x-3 equation:

=10Y^2+28Y-3

We move all terms to the left:

-(10Y^2+28Y-3)=0

We get rid of parentheses

-10Y^2-28Y+3=0

a = -10; b = -28; c = +3;
Δ = b2-4ac
Δ = -282-4·(-10)·3
Δ = 904
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:

$\sqrt{\Delta}=\sqrt{904}=\sqrt{4*226}=\sqrt{4}*\sqrt{226}=2\sqrt{226}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{226}}{2*-10}=\frac{28-2\sqrt{226}}{-20}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{226}}{2*-10}=\frac{28+2\sqrt{226}}{-20}
$