3x^2+6x^2=18

Solution for 3x^2+6x^2=18 equation:

3x^2+6x^2=18

We move all terms to the left:

3x^2+6x^2-(18)=0

We add all the numbers together, and all the variables

9x^2-18=0

a = 9; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·9·(-18)
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{2}}{2*9}=\frac{0-18\sqrt{2}}{18}
=-\frac{18\sqrt{2}}{18}
=-\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{2}}{2*9}=\frac{0+18\sqrt{2}}{18}
=\frac{18\sqrt{2}}{18}
=\sqrt{2}
$