282+b2=532

Solution for 282+b2=532 equation:

282+b2=532

We move all terms to the left:

282+b2-(532)=0

We add all the numbers together, and all the variables

b^2-250=0

a = 1; b = 0; c = -250;
Δ = b2-4ac
Δ = 02-4·1·(-250)
Δ = 1000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1000}=\sqrt{100*10}=\sqrt{100}*\sqrt{10}=10\sqrt{10}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{10}}{2*1}=\frac{0-10\sqrt{10}}{2}
=-\frac{10\sqrt{10}}{2}
=-5\sqrt{10}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{10}}{2*1}=\frac{0+10\sqrt{10}}{2}
=\frac{10\sqrt{10}}{2}
=5\sqrt{10}
$