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

The solutions to the equation x^2+3x=2(x-1) are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2

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

The zeros of x^2+3x=2(x-1) are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2

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

x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

Solution steps

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

Expand

2 (x - 1) : 2 x - 2

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

Move

2

to the left side

x^{2} + 3 x + 2 = 2 x

Move

2 x

to the left side

x^{2} + x + 2 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot 2 : 7 i

x_{1, 2} = \frac{- 1 \pm 7 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 7 i}{2 \cdot 1}, x_{2} = \frac{- 1 - 7 i}{2 \cdot 1}

x = \frac{- 1 + 7 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{7}{2}

x = \frac{- 1 - 7 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

0 = y (4 - y)

x, f (h (x)) = x^{2} + 7

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

2 x^{2} + 5 x - 8 = 0

6 x^{2} - 17 x = - 12

What are the solutions to the equation x^2+3x=2(x-1) ?
The solutions to the equation x^2+3x=2(x-1) are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2
Find the zeros of x^2+3x=2(x-1)
The zeros of x^2+3x=2(x-1) are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2