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a^2+16=100
We move all terms to the left:
a^2+16-(100)=0
We add all the numbers together, and all the variables
a^2-84=0
a = 1; b = 0; c = -84;
Δ = b2-4ac
Δ = 02-4·1·(-84)
Δ = 336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*1}=\frac{0-4\sqrt{21}}{2} =-\frac{4\sqrt{21}}{2} =-2\sqrt{21} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*1}=\frac{0+4\sqrt{21}}{2} =\frac{4\sqrt{21}}{2} =2\sqrt{21} $
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