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n2=98
We move all terms to the left:
n2-(98)=0
We add all the numbers together, and all the variables
n^2-98=0
a = 1; b = 0; c = -98;
Δ = b2-4ac
Δ = 02-4·1·(-98)
Δ = 392
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{2}}{2*1}=\frac{0-14\sqrt{2}}{2} =-\frac{14\sqrt{2}}{2} =-7\sqrt{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{2}}{2*1}=\frac{0+14\sqrt{2}}{2} =\frac{14\sqrt{2}}{2} =7\sqrt{2} $
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