-2y^2-10y-46=-y^2+4y-40

Solution for -2y^2-10y-46=-y^2+4y-40 equation:

-2y^2-10y-46=-y^2+4y-40

We move all terms to the left:

-2y^2-10y-46-(-y^2+4y-40)=0

We get rid of parentheses

-2y^2+y^2-4y-10y+40-46=0

We add all the numbers together, and all the variables

-1y^2-14y-6=0

a = -1; b = -14; c = -6;
Δ = b2-4ac
Δ = -142-4·(-1)·(-6)
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{43}}{2*-1}=\frac{14-2\sqrt{43}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{43}}{2*-1}=\frac{14+2\sqrt{43}}{-2}
$