(6x)^2+x^2=289

Solution for (6x)^2+x^2=289 equation:

(6x)^2+x^2=289

We move all terms to the left:

(6x)^2+x^2-(289)=0

We add all the numbers together, and all the variables

7x^2-289=0

a = 7; b = 0; c = -289;
Δ = b2-4ac
Δ = 02-4·7·(-289)
Δ = 8092
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:

$\sqrt{\Delta}=\sqrt{8092}=\sqrt{1156*7}=\sqrt{1156}*\sqrt{7}=34\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-34\sqrt{7}}{2*7}=\frac{0-34\sqrt{7}}{14}
=-\frac{34\sqrt{7}}{14}
=-\frac{17\sqrt{7}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+34\sqrt{7}}{2*7}=\frac{0+34\sqrt{7}}{14}
=\frac{34\sqrt{7}}{14}
=\frac{17\sqrt{7}}{7}
$