(3x(3x+x))=(6(6+2))

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

(3x(3x+x))=(6(6+2))

We move all terms to the left:

(3x(3x+x))-((6(6+2)))=0

We add all the numbers together, and all the variables

(3x(+4x))-((68))=0

We add all the numbers together, and all the variables

(3x(+4x))-68=0


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

3x(+4x)

We multiply parentheses

12x^2

Back to the equation:

+(12x^2)


a = 12; b = 0; c = -68;
Δ = b2-4ac
Δ = 02-4·12·(-68)
Δ = 3264
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:

$\sqrt{\Delta}=\sqrt{3264}=\sqrt{64*51}=\sqrt{64}*\sqrt{51}=8\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{51}}{2*12}=\frac{0-8\sqrt{51}}{24}
=-\frac{8\sqrt{51}}{24}
=-\frac{\sqrt{51}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{51}}{2*12}=\frac{0+8\sqrt{51}}{24}
=\frac{8\sqrt{51}}{24}
=\frac{\sqrt{51}}{3}
$