5n^2-4n-20=1

Solution for 5n^2-4n-20=1 equation:

5n^2-4n-20=1

We move all terms to the left:

5n^2-4n-20-(1)=0

We add all the numbers together, and all the variables

5n^2-4n-21=0

a = 5; b = -4; c = -21;
Δ = b2-4ac
Δ = -42-4·5·(-21)
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{109}}{2*5}=\frac{4-2\sqrt{109}}{10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{109}}{2*5}=\frac{4+2\sqrt{109}}{10}
$