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a^2+100=196
We move all terms to the left:
a^2+100-(196)=0
We add all the numbers together, and all the variables
a^2-96=0
a = 1; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·1·(-96)
Δ = 384
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*1}=\frac{0-8\sqrt{6}}{2} =-\frac{8\sqrt{6}}{2} =-4\sqrt{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*1}=\frac{0+8\sqrt{6}}{2} =\frac{8\sqrt{6}}{2} =4\sqrt{6} $
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