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3c^2+4c+8=-6c
We move all terms to the left:
3c^2+4c+8-(-6c)=0
We get rid of parentheses
3c^2+4c+6c+8=0
We add all the numbers together, and all the variables
3c^2+10c+8=0
a = 3; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·3·8
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$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*3}=\frac{-12}{6} =-2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*3}=\frac{-8}{6} =-1+1/3 $
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