(2l)^2+l^2=17^2

Solution for (2l)^2+l^2=17^2 equation:

(2l)^2+l^2=17^2

We move all terms to the left:

(2l)^2+l^2-(17^2)=0

We add all the numbers together, and all the variables

3l^2-289=0

a = 3; b = 0; c = -289;
Δ = b2-4ac
Δ = 02-4·3·(-289)
Δ = 3468
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3468}=\sqrt{1156*3}=\sqrt{1156}*\sqrt{3}=34\sqrt{3}$
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-34\sqrt{3}}{2*3}=\frac{0-34\sqrt{3}}{6}
=-\frac{34\sqrt{3}}{6}
=-\frac{17\sqrt{3}}{3}
$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+34\sqrt{3}}{2*3}=\frac{0+34\sqrt{3}}{6}
=\frac{34\sqrt{3}}{6}
=\frac{17\sqrt{3}}{3}
$