What are the solutions to the equation x^2+6x+3=-x(x-1) ?

The solutions to the equation x^2+6x+3=-x(x-1) are x=-1,x=-3/2

Find the zeros of x^2+6x+3=-x(x-1)

The zeros of x^2+6x+3=-x(x-1) are x=-1,x=-3/2

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x^{2} + 6 x + 3 = - x (x - 1)

x = - 1, x = - \frac{3}{2}

Decimal

x = - 1, x = - 1.5

Solution steps

x^{2} + 6 x + 3 = - x (x - 1)

Expand

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

x^{2} + 6 x + 3 = - x^{2} + x

Move

x

to the left side

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

Move

x^{2}

to the left side

2 x^{2} + 5 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 3 = 1

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

Separate the solutions

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

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

x = \frac{- 5 - 1}{2 \cdot 2} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = - 1, x = - \frac{3}{2}

(\frac{16}{9}) x^{2} = 64

x, 3 x^{2} - 2 + 2 s in x = 0

3 x^{2} - 16 x + 3 = 0

5 r^{2} - 7 r + 4 = 6

x, 6 x^{2} - 18 x = 0

What are the solutions to the equation x^2+6x+3=-x(x-1) ?
The solutions to the equation x^2+6x+3=-x(x-1) are x=-1,x=-3/2
Find the zeros of x^2+6x+3=-x(x-1)
The zeros of x^2+6x+3=-x(x-1) are x=-1,x=-3/2