5=4n^2-n-5

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

5=4n^2-n-5

We move all terms to the left:

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

We get rid of parentheses

-4n^2+n+5+5=0

We add all the numbers together, and all the variables

-4n^2+n+10=0

a = -4; b = 1; c = +10;
Δ = b2-4ac
Δ = 12-4·(-4)·10
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{161}}{2*-4}=\frac{-1-\sqrt{161}}{-8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{161}}{2*-4}=\frac{-1+\sqrt{161}}{-8}
$