3x(4x+4)=6(x^2+1)

Solution for 3x(4x+4)=6(x^2+1) equation:

3x(4x+4)=6(x^2+1)

We move all terms to the left:

3x(4x+4)-(6(x^2+1))=0

We multiply parentheses

12x^2+12x-(6(x^2+1))=0


We calculate terms in parentheses: -(6(x^2+1)), so:

6(x^2+1)

We multiply parentheses

6x^2+6

Back to the equation:

-(6x^2+6)


We get rid of parentheses

12x^2-6x^2+12x-6=0

We add all the numbers together, and all the variables

6x^2+12x-6=0

a = 6; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·6·(-6)
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{2}}{2*6}=\frac{-12-12\sqrt{2}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{2}}{2*6}=\frac{-12+12\sqrt{2}}{12}
$