x^2+(3x)^2=490

Solution for x^2+(3x)^2=490 equation:

x^2+(3x)^2=490

We move all terms to the left:

x^2+(3x)^2-(490)=0

We add all the numbers together, and all the variables

4x^2-490=0

a = 4; b = 0; c = -490;
Δ = b2-4ac
Δ = 02-4·4·(-490)
Δ = 7840
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:

$\sqrt{\Delta}=\sqrt{7840}=\sqrt{784*10}=\sqrt{784}*\sqrt{10}=28\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28\sqrt{10}}{2*4}=\frac{0-28\sqrt{10}}{8}
=-\frac{28\sqrt{10}}{8}
=-\frac{7\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28\sqrt{10}}{2*4}=\frac{0+28\sqrt{10}}{8}
=\frac{28\sqrt{10}}{8}
=\frac{7\sqrt{10}}{2}
$