n(1/5)-8=95

Solution for n(1/5)-8=95 equation:

n(1/5)-8=95

We move all terms to the left:

n(1/5)-8-(95)=0

We add all the numbers together, and all the variables

n(+1/5)-8-95=0

We add all the numbers together, and all the variables

n(+1/5)-103=0

We multiply parentheses

n^2-103=0

a = 1; b = 0; c = -103;
Δ = b2-4ac
Δ = 02-4·1·(-103)
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{-(0)-2\sqrt{103}}{2*1}=\frac{0-2\sqrt{103}}{2}
=-\frac{2\sqrt{103}}{2}
=-\sqrt{103}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{103}}{2*1}=\frac{0+2\sqrt{103}}{2}
=\frac{2\sqrt{103}}{2}
=\sqrt{103}
$