n2-2n-48=0

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Solution for n2-2n-48=0 equation:



n2-2n-48=0
We add all the numbers together, and all the variables
n^2-2n-48=0
a = 1; b = -2; c = -48;
Δ = b2-4ac
Δ = -22-4·1·(-48)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*1}=\frac{-12}{2} =-6 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*1}=\frac{16}{2} =8 $

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