(-y^2+2y-5)+(7y^2+3y+4)=0

Solution for (-y^2+2y-5)+(7y^2+3y+4)=0 equation:

(-y^2+2y-5)+(7y^2+3y+4)=0

We get rid of parentheses

-y^2+7y^2+2y+3y-5+4=0

We add all the numbers together, and all the variables

6y^2+5y-1=0

a = 6; b = 5; c = -1;
Δ = b2-4ac
Δ = 52-4·6·(-1)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*6}=\frac{-12}{12}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*6}=\frac{2}{12}
=1/6
$