C=(n+2)(n+5)

Solution for C=(n+2)(n+5) equation:

=(C+2)(C+5)

We move all terms to the left:

-((C+2)(C+5))=0

We multiply parentheses ..

-((+C^2+5C+2C+10))=0


We calculate terms in parentheses: -((+C^2+5C+2C+10)), so:

(+C^2+5C+2C+10)

We get rid of parentheses

C^2+5C+2C+10

We add all the numbers together, and all the variables

C^2+7C+10

Back to the equation:

-(C^2+7C+10)


We get rid of parentheses

-C^2-7C-10=0

We add all the numbers together, and all the variables

-1C^2-7C-10=0

a = -1; b = -7; c = -10;
Δ = b2-4ac
Δ = -72-4·(-1)·(-10)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$C_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$C_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*-1}=\frac{4}{-2}
=-2
$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*-1}=\frac{10}{-2}
=-5
$