(3y-4)(8y+1)=11(8y+1)

Solution for (3y-4)(8y+1)=11(8y+1) equation:

(3y-4)(8y+1)=11(8y+1)

We move all terms to the left:

(3y-4)(8y+1)-(11(8y+1))=0

We multiply parentheses ..

(+24y^2+3y-32y-4)-(11(8y+1))=0


We calculate terms in parentheses: -(11(8y+1)), so:

11(8y+1)

We multiply parentheses

88y+11

Back to the equation:

-(88y+11)


We get rid of parentheses

24y^2+3y-32y-88y-4-11=0

We add all the numbers together, and all the variables

24y^2-117y-15=0

a = 24; b = -117; c = -15;
Δ = b2-4ac
Δ = -1172-4·24·(-15)
Δ = 15129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15129}=123$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-117)-123}{2*24}=\frac{-6}{48}
=-1/8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-117)+123}{2*24}=\frac{240}{48}
=5
$