(21x)^2+(9x)^2=29^2

Solution for (21x)^2+(9x)^2=29^2 equation:

(21x)^2+(9x)^2=29^2

We move all terms to the left:

(21x)^2+(9x)^2-(29^2)=0

We add all the numbers together, and all the variables

30x^2-841=0

a = 30; b = 0; c = -841;
Δ = b2-4ac
Δ = 02-4·30·(-841)
Δ = 100920
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:

$\sqrt{\Delta}=\sqrt{100920}=\sqrt{3364*30}=\sqrt{3364}*\sqrt{30}=58\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-58\sqrt{30}}{2*30}=\frac{0-58\sqrt{30}}{60}
=-\frac{58\sqrt{30}}{60}
=-\frac{29\sqrt{30}}{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+58\sqrt{30}}{2*30}=\frac{0+58\sqrt{30}}{60}
=\frac{58\sqrt{30}}{60}
=\frac{29\sqrt{30}}{30}
$