4(x^2-1)=12(3x+3)

Solution for 4(x^2-1)=12(3x+3) equation:

4(x^2-1)=12(3x+3)

We move all terms to the left:

4(x^2-1)-(12(3x+3))=0

We multiply parentheses

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


We calculate terms in parentheses: -(12(3x+3)), so:

12(3x+3)

We multiply parentheses

36x+36

Back to the equation:

-(36x+36)


We get rid of parentheses

4x^2-36x-36-4=0

We add all the numbers together, and all the variables

4x^2-36x-40=0

a = 4; b = -36; c = -40;
Δ = b2-4ac
Δ = -362-4·4·(-40)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-44}{2*4}=\frac{-8}{8}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+44}{2*4}=\frac{80}{8}
=10
$