(2+c)2=(c+5)c

Solution for (2+c)2=(c+5)c equation:

(2+c)2=(c+5)c

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2c-((c+5)c)+4=0


We calculate terms in parentheses: -((c+5)c), so:

(c+5)c

We multiply parentheses

c^2+5c

Back to the equation:

-(c^2+5c)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1c^2-3c+4=0

a = -1; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-1)·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*-1}=\frac{-2}{-2}
=1
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*-1}=\frac{8}{-2}
=-4
$