(-y^2+12)/4=y

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

(-y^2+12)/4=y

We move all terms to the left:

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

We add all the numbers together, and all the variables

(-y^2+12)/4-1y=0

We multiply all the terms by the denominator


(-y^2+12)-1y*4=0

Wy multiply elements

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

We get rid of parentheses

-y^2-4y+12=0

We add all the numbers together, and all the variables

-1y^2-4y+12=0

a = -1; b = -4; c = +12;
Δ = b2-4ac
Δ = -42-4·(-1)·12
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-1}=\frac{-4}{-2}
=+2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-1}=\frac{12}{-2}
=-6
$