(2x^2)+4x^2=42

Solution for (2x^2)+4x^2=42 equation:

(2x^2)+4x^2=42

We move all terms to the left:

(2x^2)+4x^2-(42)=0

We add all the numbers together, and all the variables

6x^2-42=0

a = 6; b = 0; c = -42;
Δ = b2-4ac
Δ = 02-4·6·(-42)
Δ = 1008
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:

$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{7}}{2*6}=\frac{0-12\sqrt{7}}{12}
=-\frac{12\sqrt{7}}{12}
=-\sqrt{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{7}}{2*6}=\frac{0+12\sqrt{7}}{12}
=\frac{12\sqrt{7}}{12}
=\sqrt{7}
$