n2+10n-39=0

Solution for n2+10n-39=0 equation:

n2+10n-39=0

We add all the numbers together, and all the variables

n^2+10n-39=0

a = 1; b = 10; c = -39;
Δ = b2-4ac
Δ = 102-4·1·(-39)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-16}{2*1}=\frac{-26}{2}
=-13
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+16}{2*1}=\frac{6}{2}
=3
$