17^2=(3x)^2+(4x)^2

Solution for 17^2=(3x)^2+(4x)^2 equation:

17^2=(3x)^2+(4x)^2

We move all terms to the left:

17^2-((3x)^2+(4x)^2)=0

We add all the numbers together, and all the variables

-(3x^2+4x^2)+289=0

We get rid of parentheses

-3x^2-4x^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}
$