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3c^2+2c=1
We move all terms to the left:
3c^2+2c-(1)=0
a = 3; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·3·(-1)
Δ = 16
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{16}=4$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*3}=\frac{-6}{6} =-1 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*3}=\frac{2}{6} =1/3 $
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