3x^2+6x^2=12

Solution for 3x^2+6x^2=12 equation:

3x^2+6x^2=12

We move all terms to the left:

3x^2+6x^2-(12)=0

We add all the numbers together, and all the variables

9x^2-12=0

a = 9; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·9·(-12)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*9}=\frac{0-12\sqrt{3}}{18}
=-\frac{12\sqrt{3}}{18}
=-\frac{2\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*9}=\frac{0+12\sqrt{3}}{18}
=\frac{12\sqrt{3}}{18}
=\frac{2\sqrt{3}}{3}
$