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n^2-10=46
We move all terms to the left:
n^2-10-(46)=0
We add all the numbers together, and all the variables
n^2-56=0
a = 1; b = 0; c = -56;
Δ = b2-4ac
Δ = 02-4·1·(-56)
Δ = 224
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{14}}{2*1}=\frac{0-4\sqrt{14}}{2} =-\frac{4\sqrt{14}}{2} =-2\sqrt{14} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{14}}{2*1}=\frac{0+4\sqrt{14}}{2} =\frac{4\sqrt{14}}{2} =2\sqrt{14} $
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