25^2=(21x)^2+(9x)^2

Solution for 25^2=(21x)^2+(9x)^2 equation:

25^2=(21x)^2+(9x)^2

We move all terms to the left:

25^2-((21x)^2+(9x)^2)=0

We add all the numbers together, and all the variables

-(21x^2+9x^2)+625=0

We get rid of parentheses

-21x^2-9x^2+625=0

We add all the numbers together, and all the variables

-30x^2+625=0

a = -30; b = 0; c = +625;
Δ = b2-4ac
Δ = 02-4·(-30)·625
Δ = 75000
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:

$\sqrt{\Delta}=\sqrt{75000}=\sqrt{2500*30}=\sqrt{2500}*\sqrt{30}=50\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-50\sqrt{30}}{2*-30}=\frac{0-50\sqrt{30}}{-60}
=-\frac{50\sqrt{30}}{-60}
=-\frac{5\sqrt{30}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+50\sqrt{30}}{2*-30}=\frac{0+50\sqrt{30}}{-60}
=\frac{50\sqrt{30}}{-60}
=\frac{5\sqrt{30}}{-6}
$