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2^2+b^2=5^2
We move all terms to the left:
2^2+b^2-(5^2)=0
We add all the numbers together, and all the variables
b^2-21=0
a = 1; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·1·(-21)
Δ = 84
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{21}}{2*1}=\frac{0-2\sqrt{21}}{2} =-\frac{2\sqrt{21}}{2} =-\sqrt{21} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{21}}{2*1}=\frac{0+2\sqrt{21}}{2} =\frac{2\sqrt{21}}{2} =\sqrt{21} $
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