45-(2c+3)=4(c+7)c

Solution for 45-(2c+3)=4(c+7)c equation:

45-(2c+3)=4(c+7)c

We move all terms to the left:

45-(2c+3)-(4(c+7)c)=0

We get rid of parentheses

-2c-(4(c+7)c)-3+45=0


We calculate terms in parentheses: -(4(c+7)c), so:

4(c+7)c

We multiply parentheses

4c^2+28c

Back to the equation:

-(4c^2+28c)


We add all the numbers together, and all the variables

-2c-(4c^2+28c)+42=0

We get rid of parentheses

-4c^2-2c-28c+42=0

We add all the numbers together, and all the variables

-4c^2-30c+42=0

a = -4; b = -30; c = +42;
Δ = b2-4ac
Δ = -302-4·(-4)·42
Δ = 1572
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1572}=\sqrt{4*393}=\sqrt{4}*\sqrt{393}=2\sqrt{393}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{393}}{2*-4}=\frac{30-2\sqrt{393}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{393}}{2*-4}=\frac{30+2\sqrt{393}}{-8}
$