32-(3c+4)=2(c+5)c

Solution for 32-(3c+4)=2(c+5)c equation:

32-(3c+4)=2(c+5)c

We move all terms to the left:

32-(3c+4)-(2(c+5)c)=0

We get rid of parentheses

-3c-(2(c+5)c)-4+32=0


We calculate terms in parentheses: -(2(c+5)c), so:

2(c+5)c

We multiply parentheses

2c^2+10c

Back to the equation:

-(2c^2+10c)


We add all the numbers together, and all the variables

-3c-(2c^2+10c)+28=0

We get rid of parentheses

-2c^2-3c-10c+28=0

We add all the numbers together, and all the variables

-2c^2-13c+28=0

a = -2; b = -13; c = +28;
Δ = b2-4ac
Δ = -132-4·(-2)·28
Δ = 393
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{-(-13)-\sqrt{393}}{2*-2}=\frac{13-\sqrt{393}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{393}}{2*-2}=\frac{13+\sqrt{393}}{-4}
$