24-(3c+4)=2(c+4)c

Solution for 24-(3c+4)=2(c+4)c equation:

24-(3c+4)=2(c+4)c

We move all terms to the left:

24-(3c+4)-(2(c+4)c)=0

We get rid of parentheses

-3c-(2(c+4)c)-4+24=0


We calculate terms in parentheses: -(2(c+4)c), so:

2(c+4)c

We multiply parentheses

2c^2+8c

Back to the equation:

-(2c^2+8c)


We add all the numbers together, and all the variables

-3c-(2c^2+8c)+20=0

We get rid of parentheses

-2c^2-3c-8c+20=0

We add all the numbers together, and all the variables

-2c^2-11c+20=0

a = -2; b = -11; c = +20;
Δ = b2-4ac
Δ = -112-4·(-2)·20
Δ = 281
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{-(-11)-\sqrt{281}}{2*-2}=\frac{11-\sqrt{281}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{281}}{2*-2}=\frac{11+\sqrt{281}}{-4}
$