42^2+18^2=x^2

Solution for 42^2+18^2=x^2 equation:

42^2+18^2=x^2

We move all terms to the left:

42^2+18^2-(x^2)=0

We add all the numbers together, and all the variables

-1x^2+2088=0

a = -1; b = 0; c = +2088;
Δ = b2-4ac
Δ = 02-4·(-1)·2088
Δ = 8352
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:

$\sqrt{\Delta}=\sqrt{8352}=\sqrt{144*58}=\sqrt{144}*\sqrt{58}=12\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{58}}{2*-1}=\frac{0-12\sqrt{58}}{-2}
=-\frac{12\sqrt{58}}{-2}
=-\frac{6\sqrt{58}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{58}}{2*-1}=\frac{0+12\sqrt{58}}{-2}
=\frac{12\sqrt{58}}{-2}
=\frac{6\sqrt{58}}{-1}
$