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

The solutions to the equation x^2-x(3x+1)=3 are x=-1/4-i(sqrt(23))/4 ,x=-1/4+i(sqrt(23))/4

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

The zeros of x^2-x(3x+1)=3 are x=-1/4-i(sqrt(23))/4 ,x=-1/4+i(sqrt(23))/4

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x^{2} - x (3 x + 1) = 3

x = - \frac{1}{4} - i \frac{23}{4}, x = - \frac{1}{4} + i \frac{23}{4}

Solution steps

x^{2} - x (3 x + 1) = 3

Expand

x^{2} - x (3 x + 1) : - 2 x^{2} - x

- 2 x^{2} - x = 3

Move

3

to the left side

- 2 x^{2} - x - 3 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 (- 2) (- 3) : 23 i

x_{1, 2} = \frac{- ( - 1 ) \pm 23 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 23 i}{2 ( - 2 )}, x_{2} = \frac{- ( - 1 ) - 23 i}{2 ( - 2 )}

x = \frac{- ( - 1 ) + 23 i}{2 ( - 2 )} : - \frac{1}{4} - i \frac{23}{4}

x = \frac{- ( - 1 ) - 23 i}{2 ( - 2 )} : - \frac{1}{4} + i \frac{23}{4}

The solutions to the quadratic equation are:

x = - \frac{1}{4} - i \frac{23}{4}, x = - \frac{1}{4} + i \frac{23}{4}

6 x^{2} - 72 x + 162 = 0

3 y^{2} - 4 = 0

3 x^{2} - 23 x + 1 = 0

- 12 x + 9 x^{2} + 4 = 0

7 x^{2} - 35 x = 0

What are the solutions to the equation x^2-x(3x+1)=3 ?
The solutions to the equation x^2-x(3x+1)=3 are x=-1/4-i(sqrt(23))/4 ,x=-1/4+i(sqrt(23))/4
Find the zeros of x^2-x(3x+1)=3
The zeros of x^2-x(3x+1)=3 are x=-1/4-i(sqrt(23))/4 ,x=-1/4+i(sqrt(23))/4