y(4y+5)=35+8y

Solution for y(4y+5)=35+8y equation:

y(4y+5)=35+8y

We move all terms to the left:

y(4y+5)-(35+8y)=0

We add all the numbers together, and all the variables

y(4y+5)-(8y+35)=0

We multiply parentheses

4y^2+5y-(8y+35)=0

We get rid of parentheses

4y^2+5y-8y-35=0

We add all the numbers together, and all the variables

4y^2-3y-35=0

a = 4; b = -3; c = -35;
Δ = b2-4ac
Δ = -32-4·4·(-35)
Δ = 569
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{569}}{2*4}=\frac{3-\sqrt{569}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{569}}{2*4}=\frac{3+\sqrt{569}}{8}
$