n^2-14n-48=0

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Solution for n^2-14n-48=0 equation:



n^2-14n-48=0
a = 1; b = -14; c = -48;
Δ = b2-4ac
Δ = -142-4·1·(-48)
Δ = 388
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:
$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{97}}{2*1}=\frac{14-2\sqrt{97}}{2} $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{97}}{2*1}=\frac{14+2\sqrt{97}}{2} $

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