What are the solutions to the equation (x-3)x+1=x^2-5x(x+1) ?

The solutions to the equation (x-3)x+1=x^2-5x(x+1) are x=-1/5+i 2/5 ,x=-1/5-i 2/5

Find the zeros of (x-3)x+1=x^2-5x(x+1)

The zeros of (x-3)x+1=x^2-5x(x+1) are x=-1/5+i 2/5 ,x=-1/5-i 2/5

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(x - 3) x + 1 = x^{2} - 5 x (x + 1)

x = - \frac{1}{5} + i \frac{2}{5}, x = - \frac{1}{5} - i \frac{2}{5}

Solution steps

(x - 3) x + 1 = x^{2} - 5 x (x + 1)

Expand

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

Expand

x^{2} - 5 x (x + 1) : - 4 x^{2} - 5 x

x^{2} - 3 x + 1 = - 4 x^{2} - 5 x

Move

5 x

to the left side

x^{2} + 2 x + 1 = - 4 x^{2}

Move

4 x^{2}

to the left side

5 x^{2} + 2 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

Simplify

2^{2} - 4 \cdot 5 \cdot 1 : 4 i

x_{1, 2} = \frac{- 2 \pm 4 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 2 + 4 i}{2 \cdot 5}, x_{2} = \frac{- 2 - 4 i}{2 \cdot 5}

x = \frac{- 2 + 4 i}{2 \cdot 5} : - \frac{1}{5} + i \frac{2}{5}

x = \frac{- 2 - 4 i}{2 \cdot 5} : - \frac{1}{5} - i \frac{2}{5}

The solutions to the quadratic equation are:

x = - \frac{1}{5} + i \frac{2}{5}, x = - \frac{1}{5} - i \frac{2}{5}

\frac{x ^{2}}{4} + \frac{1}{10} = \frac{x ^{2}}{5} + \frac{x}{2}

2 x^{2} + 15 x - 7 = 2 x

(x + 4) (x - 3) = 8

(3 x)^{2} - 4 (x + 5) (x) = 0

x, y^{3} - x y + x^{2} = 1

What are the solutions to the equation (x-3)x+1=x^2-5x(x+1) ?
The solutions to the equation (x-3)x+1=x^2-5x(x+1) are x=-1/5+i 2/5 ,x=-1/5-i 2/5
Find the zeros of (x-3)x+1=x^2-5x(x+1)
The zeros of (x-3)x+1=x^2-5x(x+1) are x=-1/5+i 2/5 ,x=-1/5-i 2/5