What are the solutions to the equation 2c(2c+1)=c(c-3)-3 ?

The solutions to the equation 2c(2c+1)=c(c-3)-3 are c=-5/6-i(sqrt(11))/6 ,c=-5/6+i(sqrt(11))/6

Find the zeros of 2c(2c+1)=c(c-3)-3

The zeros of 2c(2c+1)=c(c-3)-3 are c=-5/6-i(sqrt(11))/6 ,c=-5/6+i(sqrt(11))/6

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2 c (2 c + 1) = c (c - 3) - 3

c = - \frac{5}{6} - i \frac{11}{6}, c = - \frac{5}{6} + i \frac{11}{6}

Solution steps

2 c (2 c + 1) = c (c - 3) - 3

Expand

2 c (2 c + 1) : 4 c^{2} + 2 c

Expand

c (c - 3) - 3 : c^{2} - 3 c - 3

4 c^{2} + 2 c = c^{2} - 3 c - 3

Switch sides

c^{2} - 3 c - 3 = 4 c^{2} + 2 c

Move

2 c

to the left side

c^{2} - 5 c - 3 = 4 c^{2}

Move

4 c^{2}

to the left side

- 3 c^{2} - 5 c - 3 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 3 ) ( - 3 )}{2 ( - 3 )}

Simplify

(- 5)^{2} - 4 (- 3) (- 3) : 11 i

c_{1, 2} = \frac{- ( - 5 ) \pm 11 i}{2 ( - 3 )}

Separate the solutions

c_{1} = \frac{- ( - 5 ) + 11 i}{2 ( - 3 )}, c_{2} = \frac{- ( - 5 ) - 11 i}{2 ( - 3 )}

c = \frac{- ( - 5 ) + 11 i}{2 ( - 3 )} : - \frac{5}{6} - i \frac{11}{6}

c = \frac{- ( - 5 ) - 11 i}{2 ( - 3 )} : - \frac{5}{6} + i \frac{11}{6}

The solutions to the quadratic equation are:

c = - \frac{5}{6} - i \frac{11}{6}, c = - \frac{5}{6} + i \frac{11}{6}

4 x^{2} = 24 - 2 x^{2}

x^{2} + 18 x + 42 = 0

3 x^{2} - 24 = - 5

(2 v + 6)^{2} = 9

z^{2} + 6 z + 4 = 0

What are the solutions to the equation 2c(2c+1)=c(c-3)-3 ?
The solutions to the equation 2c(2c+1)=c(c-3)-3 are c=-5/6-i(sqrt(11))/6 ,c=-5/6+i(sqrt(11))/6
Find the zeros of 2c(2c+1)=c(c-3)-3
The zeros of 2c(2c+1)=c(c-3)-3 are c=-5/6-i(sqrt(11))/6 ,c=-5/6+i(sqrt(11))/6