7n^2-34-48=0

Solution for 7n^2-34-48=0 equation:

7n^2-34-48=0

We add all the numbers together, and all the variables

7n^2-82=0

a = 7; b = 0; c = -82;
Δ = b2-4ac
Δ = 02-4·7·(-82)
Δ = 2296
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:

$\sqrt{\Delta}=\sqrt{2296}=\sqrt{4*574}=\sqrt{4}*\sqrt{574}=2\sqrt{574}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{574}}{2*7}=\frac{0-2\sqrt{574}}{14}
=-\frac{2\sqrt{574}}{14}
=-\frac{\sqrt{574}}{7}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{574}}{2*7}=\frac{0+2\sqrt{574}}{14}
=\frac{2\sqrt{574}}{14}
=\frac{\sqrt{574}}{7}
$