6(4+c)=5(3+c)c=

Solution for 6(4+c)=5(3+c)c= equation:

6(4+c)=5(3+c)c=

We move all terms to the left:

6(4+c)-(5(3+c)c)=0

We add all the numbers together, and all the variables

6(c+4)-(5(c+3)c)=0

We multiply parentheses

6c-(5(c+3)c)+24=0


We calculate terms in parentheses: -(5(c+3)c), so:

5(c+3)c

We multiply parentheses

5c^2+15c

Back to the equation:

-(5c^2+15c)


We get rid of parentheses

-5c^2+6c-15c+24=0

We add all the numbers together, and all the variables

-5c^2-9c+24=0

a = -5; b = -9; c = +24;
Δ = b2-4ac
Δ = -92-4·(-5)·24
Δ = 561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{561}}{2*-5}=\frac{9-\sqrt{561}}{-10}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{561}}{2*-5}=\frac{9+\sqrt{561}}{-10}
$