(8y^2+6y-5)-(-2y+4)=0

Solution for (8y^2+6y-5)-(-2y+4)=0 equation:

(8y^2+6y-5)-(-2y+4)=0

We get rid of parentheses

8y^2+6y+2y-5-4=0

We add all the numbers together, and all the variables

8y^2+8y-9=0

a = 8; b = 8; c = -9;
Δ = b2-4ac
Δ = 82-4·8·(-9)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{22}}{2*8}=\frac{-8-4\sqrt{22}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{22}}{2*8}=\frac{-8+4\sqrt{22}}{16}
$