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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

y+((2y+6)/2)+((2y+6)/2)-6=0

We multiply all the terms by the denominator


y*2)+((2y+6)+((2y+6)-6*2)=0


We calculate terms in parentheses: +((2y+6)-6*2), so:

(2y+6)-6*2

We add all the numbers together, and all the variables

(2y+6)-12

We get rid of parentheses

2y+6-12

We add all the numbers together, and all the variables

2y-6

Back to the equation:

+(2y-6)


Wy multiply elements

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

We get rid of parentheses

2y^2+2y-6=0

a = 2; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·2·(-6)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*2}=\frac{-2-2\sqrt{13}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*2}=\frac{-2+2\sqrt{13}}{4}
$