5(3c2)-7c=40-2c

Solution for 5(3c2)-7c=40-2c equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

53c^2-7c-(-2c+40)=0

We get rid of parentheses

53c^2-7c+2c-40=0

We add all the numbers together, and all the variables

53c^2-5c-40=0

a = 53; b = -5; c = -40;
Δ = b2-4ac
Δ = -52-4·53·(-40)
Δ = 8505
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:

$\sqrt{\Delta}=\sqrt{8505}=\sqrt{81*105}=\sqrt{81}*\sqrt{105}=9\sqrt{105}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9\sqrt{105}}{2*53}=\frac{5-9\sqrt{105}}{106}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9\sqrt{105}}{2*53}=\frac{5+9\sqrt{105}}{106}
$