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

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

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

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

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

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

Decimal Notation

x = - 0.5, x = 3

Solution steps

2 (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

(\frac{1}{4}) (x - 3) = 5 - x

2 + 23 i

x^{2} + 8 x - 4 = 0

- \frac{1}{2} y^{2} + \frac{9}{8}

3 (25 m^{6}) (\frac{3 n ^{2}}{5})

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