2(y+2)4y=3(2y-)+8

Solution for 2(y+2)4y=3(2y-)+8 equation:

2(y+2)4y=3(2y-)+8

We move all terms to the left:

2(y+2)4y-(3(2y-)+8)=0

We add all the numbers together, and all the variables

2(y+2)4y-(3(+2y)+8)=0

We multiply parentheses

8y^2+16y-(3(+2y)+8)=0


We calculate terms in parentheses: -(3(+2y)+8), so:

3(+2y)+8

We multiply parentheses

6y+8

Back to the equation:

-(6y+8)


We get rid of parentheses

8y^2+16y-6y-8=0

We add all the numbers together, and all the variables

8y^2+10y-8=0

a = 8; b = 10; c = -8;
Δ = b2-4ac
Δ = 102-4·8·(-8)
Δ = 356
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:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{89}}{2*8}=\frac{-10-2\sqrt{89}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{89}}{2*8}=\frac{-10+2\sqrt{89}}{16}
$