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

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

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

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

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

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

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

x + 3 = 0 : x = - 3

The solutions to the quadratic equation are:

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

16 z^{2} - 9

6 (5 t - 2) = - 31 (7 - t)

9 r^{8} + 14 r^{4} v^{4} - 8 v^{8}

x + 4 = - 14

3 (x - 2) = 12

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