4(n+2)=5n^2+2

Solution for 4(n+2)=5n^2+2 equation:

4(n+2)=5n^2+2

We move all terms to the left:

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

We multiply parentheses

4n-(5n^2+2)+8=0

We get rid of parentheses

-5n^2+4n-2+8=0

We add all the numbers together, and all the variables

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

a = -5; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-5)·6
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{34}}{2*-5}=\frac{-4-2\sqrt{34}}{-10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{34}}{2*-5}=\frac{-4+2\sqrt{34}}{-10}
$