27-(c2+4)=2(c+4)+c

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Solution for 27-(c2+4)=2(c+4)+c equation:



27-(c2+4)=2(c+4)+c
We move all terms to the left:
27-(c2+4)-(2(c+4)+c)=0
We add all the numbers together, and all the variables
-(+c^2+4)-(2(c+4)+c)+27=0
We get rid of parentheses
-c^2-(2(c+4)+c)-4+27=0
We calculate terms in parentheses: -(2(c+4)+c), so:
2(c+4)+c
We add all the numbers together, and all the variables
c+2(c+4)
We multiply parentheses
c+2c+8
We add all the numbers together, and all the variables
3c+8
Back to the equation:
-(3c+8)
We add all the numbers together, and all the variables
-1c^2-(3c+8)+23=0
We get rid of parentheses
-1c^2-3c-8+23=0
We add all the numbers together, and all the variables
-1c^2-3c+15=0
a = -1; b = -3; c = +15;
Δ = b2-4ac
Δ = -32-4·(-1)·15
Δ = 69
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{-(-3)-\sqrt{69}}{2*-1}=\frac{3-\sqrt{69}}{-2} $
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{69}}{2*-1}=\frac{3+\sqrt{69}}{-2} $

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