5c(3c-2)-7c=40-2c

Solution for 5c(3c-2)-7c=40-2c equation:

5c(3c-2)-7c=40-2c

We move all terms to the left:

5c(3c-2)-7c-(40-2c)=0

We add all the numbers together, and all the variables

5c(3c-2)-7c-(-2c+40)=0

We add all the numbers together, and all the variables

-7c+5c(3c-2)-(-2c+40)=0

We multiply parentheses

15c^2-7c-10c-(-2c+40)=0

We get rid of parentheses

15c^2-7c-10c+2c-40=0

We add all the numbers together, and all the variables

15c^2-15c-40=0

a = 15; b = -15; c = -40;
Δ = b2-4ac
Δ = -152-4·15·(-40)
Δ = 2625
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:

$\sqrt{\Delta}=\sqrt{2625}=\sqrt{25*105}=\sqrt{25}*\sqrt{105}=5\sqrt{105}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5\sqrt{105}}{2*15}=\frac{15-5\sqrt{105}}{30}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5\sqrt{105}}{2*15}=\frac{15+5\sqrt{105}}{30}
$