7n2+41-54=2

Solution for 7n2+41-54=2 equation:

7n^2+41-54=2

We move all terms to the left:

7n^2+41-54-(2)=0

We add all the numbers together, and all the variables

7n^2-15=0

a = 7; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·7·(-15)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{105}}{2*7}=\frac{0-2\sqrt{105}}{14}
=-\frac{2\sqrt{105}}{14}
=-\frac{\sqrt{105}}{7}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{105}}{2*7}=\frac{0+2\sqrt{105}}{14}
=\frac{2\sqrt{105}}{14}
=\frac{\sqrt{105}}{7}
$