5n2+n-4=6

Solution for 5n2+n-4=6 equation:

5n^2+n-4=6

We move all terms to the left:

5n^2+n-4-(6)=0

We add all the numbers together, and all the variables

5n^2+n-10=0

a = 5; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·5·(-10)
Δ = 201
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{201}}{2*5}=\frac{-1-\sqrt{201}}{10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{201}}{2*5}=\frac{-1+\sqrt{201}}{10}
$