8y(y+4)=7y+38

Solution for 8y(y+4)=7y+38 equation:

8y(y+4)=7y+38

We move all terms to the left:

8y(y+4)-(7y+38)=0

We multiply parentheses

8y^2+32y-(7y+38)=0

We get rid of parentheses

8y^2+32y-7y-38=0

We add all the numbers together, and all the variables

8y^2+25y-38=0

a = 8; b = 25; c = -38;
Δ = b2-4ac
Δ = 252-4·8·(-38)
Δ = 1841
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{-(25)-\sqrt{1841}}{2*8}=\frac{-25-\sqrt{1841}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{1841}}{2*8}=\frac{-25+\sqrt{1841}}{16}
$