Y=2x^2+4x-11

Solution for Y=2x^2+4x-11 equation:

=2Y^2+4Y-11

We move all terms to the left:

-(2Y^2+4Y-11)=0

We get rid of parentheses

-2Y^2-4Y+11=0

a = -2; b = -4; c = +11;
Δ = b2-4ac
Δ = -42-4·(-2)·11
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{26}}{2*-2}=\frac{4-2\sqrt{26}}{-4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{26}}{2*-2}=\frac{4+2\sqrt{26}}{-4}
$