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

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

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

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

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

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

Decimal Notation

x = - 1, x = 1.5

Solution steps

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

Move

x (x + 1)

to the left side

(3 x + 3) (x - 1) - x (x + 1) = 0

Factor

(3 x + 3) (x - 1) - x (x + 1) : (x + 1) (2 x - 3)

(x + 1) (2 x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or 2 x - 3 = 0

Solve

x + 1 = 0 : x = - 1

Solve

2 x - 3 = 0 : x = \frac{3}{2}

The solutions to the quadratic equation are:

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

x^{2} - 4 x = 6 x - x^{2}

x^{2} = 1 5^{2} + 3 6^{2}

8 x^{2} - 2 = x^{2} + 12

d, v = \frac{π d ^{2} h}{4}

12 x^{2} + 36 x - 48 = 0

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