(6c-4)-(c-3)c=-3

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Solution for (6c-4)-(c-3)c=-3 equation:



(6c-4)-(c-3)c=-3
We move all terms to the left:
(6c-4)-(c-3)c-(-3)=0
We add all the numbers together, and all the variables
(6c-4)-(c-3)c+3=0
We multiply parentheses
-c^2+(6c-4)+3c+3=0
We get rid of parentheses
-c^2+6c+3c-4+3=0
We add all the numbers together, and all the variables
-1c^2+9c-1=0
a = -1; b = 9; c = -1;
Δ = b2-4ac
Δ = 92-4·(-1)·(-1)
Δ = 77
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{77}}{2*-1}=\frac{-9-\sqrt{77}}{-2} $
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{77}}{2*-1}=\frac{-9+\sqrt{77}}{-2} $

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