2x^2+x2=38^2

Solution for 2x^2+x2=38^2 equation:

2x^2+x2=38^2

We move all terms to the left:

2x^2+x2-(38^2)=0

We add all the numbers together, and all the variables

3x^2-1444=0

a = 3; b = 0; c = -1444;
Δ = b2-4ac
Δ = 02-4·3·(-1444)
Δ = 17328
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:

$\sqrt{\Delta}=\sqrt{17328}=\sqrt{5776*3}=\sqrt{5776}*\sqrt{3}=76\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-76\sqrt{3}}{2*3}=\frac{0-76\sqrt{3}}{6}
=-\frac{76\sqrt{3}}{6}
=-\frac{38\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+76\sqrt{3}}{2*3}=\frac{0+76\sqrt{3}}{6}
=\frac{76\sqrt{3}}{6}
=\frac{38\sqrt{3}}{3}
$