(4y^2-8y+12)/(y^2+y-6)=3

Solution for (4y^2-8y+12)/(y^2+y-6)=3 equation:

(4y^2-8y+12)/(y^2+y-6)=3

We move all terms to the left:

(4y^2-8y+12)/(y^2+y-6)-(3)=0


Domain of the equation: (y^2+y-6)!=0

We move all terms containing y to the left, all other terms to the right

y^2+y!=6

y∈R


We multiply all the terms by the denominator


(4y^2-8y+12)-3*(y^2+y-6)=0

We multiply parentheses

-3y^2+(4y^2-8y+12)-3y+18=0

We get rid of parentheses

-3y^2+4y^2-8y-3y+12+18=0

We add all the numbers together, and all the variables

y^2-11y+30=0

a = 1; b = -11; c = +30;
Δ = b2-4ac
Δ = -112-4·1·30
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2*1}=\frac{10}{2}
=5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2*1}=\frac{12}{2}
=6
$