7y(y+3)=5y+8

Solution for 7y(y+3)=5y+8 equation:

7y(y+3)=5y+8

We move all terms to the left:

7y(y+3)-(5y+8)=0

We multiply parentheses

7y^2+21y-(5y+8)=0

We get rid of parentheses

7y^2+21y-5y-8=0

We add all the numbers together, and all the variables

7y^2+16y-8=0

a = 7; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·7·(-8)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{30}}{2*7}=\frac{-16-4\sqrt{30}}{14}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{30}}{2*7}=\frac{-16+4\sqrt{30}}{14}
$