What are the solutions to the equation (3x+2)(x+1)=0 ?

The solutions to the equation (3x+2)(x+1)=0 are x=-2/3 ,x=-1

Find the zeros of (3x+2)(x+1)=0

The zeros of (3x+2)(x+1)=0 are x=-2/3 ,x=-1

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

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

Decimal Notation

x = - 0.66666 \dots, x = - 1

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

x + 1 = 0 : x = - 1

The solutions to the quadratic equation are:

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

\frac{20}{45}

2 s^{2} + 8 s + 8 = 0

5 - 3 z + 3 (3 - 5 z) = 12

\frac{k ^{2} - k - 2}{k ^{2} - 4 k - 5}

(x - 4) (x - 5) = 0

What are the solutions to the equation (3x+2)(x+1)=0 ?
The solutions to the equation (3x+2)(x+1)=0 are x=-2/3 ,x=-1
Find the zeros of (3x+2)(x+1)=0
The zeros of (3x+2)(x+1)=0 are x=-2/3 ,x=-1