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3(c+1)c=9
We move all terms to the left:
3(c+1)c-(9)=0
We multiply parentheses
3c^2+3c-9=0
a = 3; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·3·(-9)
Δ = 117
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{13}}{2*3}=\frac{-3-3\sqrt{13}}{6} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{13}}{2*3}=\frac{-3+3\sqrt{13}}{6} $
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