Y=2x^2-10x-5

Solution for Y=2x^2-10x-5 equation:

=2Y^2-10Y-5

We move all terms to the left:

-(2Y^2-10Y-5)=0

We get rid of parentheses

-2Y^2+10Y+5=0

a = -2; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·(-2)·5
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{35}}{2*-2}=\frac{-10-2\sqrt{35}}{-4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{35}}{2*-2}=\frac{-10+2\sqrt{35}}{-4}
$