(2c-6)3c-3=27

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Solution for (2c-6)3c-3=27 equation:



(2c-6)3c-3=27
We move all terms to the left:
(2c-6)3c-3-(27)=0
We add all the numbers together, and all the variables
(2c-6)3c-30=0
We multiply parentheses
6c^2-18c-30=0
a = 6; b = -18; c = -30;
Δ = b2-4ac
Δ = -182-4·6·(-30)
Δ = 1044
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:
$\sqrt{\Delta}=\sqrt{1044}=\sqrt{36*29}=\sqrt{36}*\sqrt{29}=6\sqrt{29}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{29}}{2*6}=\frac{18-6\sqrt{29}}{12} $
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{29}}{2*6}=\frac{18+6\sqrt{29}}{12} $

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