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n^2+2+10n-39=0
We add all the numbers together, and all the variables
n^2+10n-37=0
a = 1; b = 10; c = -37;
Δ = b2-4ac
Δ = 102-4·1·(-37)
Δ = 248
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{62}}{2*1}=\frac{-10-2\sqrt{62}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{62}}{2*1}=\frac{-10+2\sqrt{62}}{2} $
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