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

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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

12y^2-3y-1=0

a = 12; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·12·(-1)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{57}}{2*12}=\frac{3-\sqrt{57}}{24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*12}=\frac{3+\sqrt{57}}{24}
$