What are the solutions to the equation x(x-1)=(3x+1)(x+2)+4 ?

The solutions to the equation x(x-1)=(3x+1)(x+2)+4 are x=-1,x=-3

Find the zeros of x(x-1)=(3x+1)(x+2)+4

The zeros of x(x-1)=(3x+1)(x+2)+4 are x=-1,x=-3

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x (x - 1) = (3 x + 1) (x + 2) + 4

x = - 1, x = - 3

Solution steps

x (x - 1) = (3 x + 1) (x + 2) + 4

Expand

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

Expand

(3 x + 1) (x + 2) + 4 : 3 x^{2} + 7 x + 6

x^{2} - x = 3 x^{2} + 7 x + 6

Switch sides

3 x^{2} + 7 x + 6 = x^{2} - x

Move

x

to the left side

3 x^{2} + 8 x + 6 = x^{2}

Move

x^{2}

to the left side

2 x^{2} + 8 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

8^{2} - 4 \cdot 2 \cdot 6 = 4

x_{1, 2} = \frac{- 8 \pm 4}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 4}{2 \cdot 2}, x_{2} = \frac{- 8 - 4}{2 \cdot 2}

x = \frac{- 8 + 4}{2 \cdot 2} : - 1

x = \frac{- 8 - 4}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

x = - 1, x = - 3

x, f = p x^{2} - 4

2 x^{2} - 10 x + 15 = 0

x^{2} + 6 x - 7 = 0

x, (x - a)^{2} + y^{2} = b

x, y^{3} - 2 x^{2} = 1388

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