22^2+7^2=d^2

Solution for 22^2+7^2=d^2 equation:

22^2+7^2=d^2

We move all terms to the left:

22^2+7^2-(d^2)=0

We add all the numbers together, and all the variables

-1d^2+533=0

a = -1; b = 0; c = +533;
Δ = b2-4ac
Δ = 02-4·(-1)·533
Δ = 2132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2132}=\sqrt{4*533}=\sqrt{4}*\sqrt{533}=2\sqrt{533}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{533}}{2*-1}=\frac{0-2\sqrt{533}}{-2}
=-\frac{2\sqrt{533}}{-2}
=-\frac{\sqrt{533}}{-1}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{533}}{2*-1}=\frac{0+2\sqrt{533}}{-2}
=\frac{2\sqrt{533}}{-2}
=\frac{\sqrt{533}}{-1}
$