10c2-21c=-4c+6

Solution for 10c2-21c=-4c+6 equation:

10c^2-21c=-4c+6

We move all terms to the left:

10c^2-21c-(-4c+6)=0

We get rid of parentheses

10c^2-21c+4c-6=0

We add all the numbers together, and all the variables

10c^2-17c-6=0

a = 10; b = -17; c = -6;
Δ = b2-4ac
Δ = -172-4·10·(-6)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-23}{2*10}=\frac{-6}{20}
=-3/10
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+23}{2*10}=\frac{40}{20}
=2
$