22-(3c+4)=2(c+3)c

Solution for 22-(3c+4)=2(c+3)c equation:

22-(3c+4)=2(c+3)c

We move all terms to the left:

22-(3c+4)-(2(c+3)c)=0

We get rid of parentheses

-3c-(2(c+3)c)-4+22=0


We calculate terms in parentheses: -(2(c+3)c), so:

2(c+3)c

We multiply parentheses

2c^2+6c

Back to the equation:

-(2c^2+6c)


We add all the numbers together, and all the variables

-3c-(2c^2+6c)+18=0

We get rid of parentheses

-2c^2-3c-6c+18=0

We add all the numbers together, and all the variables

-2c^2-9c+18=0

a = -2; b = -9; c = +18;
Δ = b2-4ac
Δ = -92-4·(-2)·18
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-15}{2*-2}=\frac{-6}{-4}
=1+1/2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+15}{2*-2}=\frac{24}{-4}
=-6
$