3n(n+5)=7-n-18

Solution for 3n(n+5)=7-n-18 equation:

3n(n+5)=7-n-18

We move all terms to the left:

3n(n+5)-(7-n-18)=0

We add all the numbers together, and all the variables

3n(n+5)-(-1n-11)=0

We multiply parentheses

3n^2+15n-(-1n-11)=0

We get rid of parentheses

3n^2+15n+1n+11=0

We add all the numbers together, and all the variables

3n^2+16n+11=0

a = 3; b = 16; c = +11;
Δ = b2-4ac
Δ = 162-4·3·11
Δ = 124
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:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{31}}{2*3}=\frac{-16-2\sqrt{31}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{31}}{2*3}=\frac{-16+2\sqrt{31}}{6}
$