y^2+12=4(3+y)

Solution for y^2+12=4(3+y) equation:

y^2+12=4(3+y)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(4(y+3)), so:

4(y+3)

We multiply parentheses

4y+12

Back to the equation:

-(4y+12)


We get rid of parentheses

y^2-4y-12+12=0

We add all the numbers together, and all the variables

y^2-4y=0

a = 1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·1·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*1}=\frac{0}{2}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*1}=\frac{8}{2}
=4
$