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3c^2=2c+2c^2
We move all terms to the left:
3c^2-(2c+2c^2)=0
We get rid of parentheses
3c^2-2c^2-2c=0
We add all the numbers together, and all the variables
c^2-2c=0
a = 1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1·0
Δ = 4
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{4}=2$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*1}=\frac{0}{2} =0 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*1}=\frac{4}{2} =2 $
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