2(4y-3)2y=8y-6+2y=10y-6

Solution for 2(4y-3)2y=8y-6+2y=10y-6 equation:

2(4y-3)2y=8y-6+2y=10y-6

We move all terms to the left:

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

We add all the numbers together, and all the variables

2(4y-3)2y-(10y-6)=0

We multiply parentheses

16y^2-12y-(10y-6)=0

We get rid of parentheses

16y^2-12y-10y+6=0

We add all the numbers together, and all the variables

16y^2-22y+6=0

a = 16; b = -22; c = +6;
Δ = b2-4ac
Δ = -222-4·16·6
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-10}{2*16}=\frac{12}{32}
=3/8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+10}{2*16}=\frac{32}{32}
=1
$