1n^2+(27n)-12=0

Solution for 1n^2+(27n)-12=0 equation:

1n^2+(27n)-12=0

We add all the numbers together, and all the variables

n^2+27n-12=0

a = 1; b = 27; c = -12;
Δ = b2-4ac
Δ = 272-4·1·(-12)
Δ = 777
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-\sqrt{777}}{2*1}=\frac{-27-\sqrt{777}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{777}}{2*1}=\frac{-27+\sqrt{777}}{2}
$