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a^2=196
We move all terms to the left:
a^2-(196)=0
a = 1; b = 0; c = -196;
Δ = b2-4ac
Δ = 02-4·1·(-196)
Δ = 784
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{784}=28$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28}{2*1}=\frac{-28}{2} =-14 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28}{2*1}=\frac{28}{2} =14 $
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