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n^2+4n-68=-2
We move all terms to the left:
n^2+4n-68-(-2)=0
We add all the numbers together, and all the variables
n^2+4n-66=0
a = 1; b = 4; c = -66;
Δ = b2-4ac
Δ = 42-4·1·(-66)
Δ = 280
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{70}}{2*1}=\frac{-4-2\sqrt{70}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{70}}{2*1}=\frac{-4+2\sqrt{70}}{2} $
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