4y(2y-3)=(y+6)

Solution for 4y(2y-3)=(y+6) equation:

4y(2y-3)=(y+6)

We move all terms to the left:

4y(2y-3)-((y+6))=0

We multiply parentheses

8y^2-12y-((y+6))=0


We calculate terms in parentheses: -((y+6)), so:

(y+6)

We get rid of parentheses

y+6

Back to the equation:

-(y+6)


We get rid of parentheses

8y^2-12y-y-6=0

We add all the numbers together, and all the variables

8y^2-13y-6=0

a = 8; b = -13; c = -6;
Δ = b2-4ac
Δ = -132-4·8·(-6)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-19}{2*8}=\frac{-6}{16}
=-3/8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+19}{2*8}=\frac{32}{16}
=2
$