n2+14n-51=3

Solution for n2+14n-51=3 equation:

n2+14n-51=3

We move all terms to the left:

n2+14n-51-(3)=0

We add all the numbers together, and all the variables

n^2+14n-54=0

a = 1; b = 14; c = -54;
Δ = b2-4ac
Δ = 142-4·1·(-54)
Δ = 412
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:

$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{103}}{2*1}=\frac{-14-2\sqrt{103}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{103}}{2*1}=\frac{-14+2\sqrt{103}}{2}
$