(4y^(2)+8y-7)/(2y+2)=y

Solution for (4y^(2)+8y-7)/(2y+2)=y equation:

(4y^(2)+8y-7)/(2y+2)=y

We move all terms to the left:

(4y^(2)+8y-7)/(2y+2)-(y)=0


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

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

2y!=-2

y!=-2/2

y!=-1

y∈R


We add all the numbers together, and all the variables

-1y+(4y^2+8y-7)/(2y+2)=0

We multiply all the terms by the denominator


-1y*(2y+2)+(4y^2+8y-7)=0

We multiply parentheses

-2y^2-2y+(4y^2+8y-7)=0

We get rid of parentheses

-2y^2+4y^2-2y+8y-7=0

We add all the numbers together, and all the variables

2y^2+6y-7=0

a = 2; b = 6; c = -7;
Δ = b2-4ac
Δ = 62-4·2·(-7)
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{23}}{2*2}=\frac{-6-2\sqrt{23}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{23}}{2*2}=\frac{-6+2\sqrt{23}}{4}
$