2(5c+2)2c=3(2c+3)+7

Solution for 2(5c+2)2c=3(2c+3)+7 equation:

2(5c+2)2c=3(2c+3)+7

We move all terms to the left:

2(5c+2)2c-(3(2c+3)+7)=0

We multiply parentheses

20c^2+8c-(3(2c+3)+7)=0


We calculate terms in parentheses: -(3(2c+3)+7), so:

3(2c+3)+7

We multiply parentheses

6c+9+7

We add all the numbers together, and all the variables

6c+16

Back to the equation:

-(6c+16)


We get rid of parentheses

20c^2+8c-6c-16=0

We add all the numbers together, and all the variables

20c^2+2c-16=0

a = 20; b = 2; c = -16;
Δ = b2-4ac
Δ = 22-4·20·(-16)
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{321}}{2*20}=\frac{-2-2\sqrt{321}}{40}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{321}}{2*20}=\frac{-2+2\sqrt{321}}{40}
$