What are the solutions to the equation (1-2x)(x+12)=0 ?

The solutions to the equation (1-2x)(x+12)=0 are x= 1/2 ,x=-12

Find the zeros of (1-2x)(x+12)=0

The zeros of (1-2x)(x+12)=0 are x= 1/2 ,x=-12

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(1 - 2 x) (x + 12) = 0

x = \frac{1}{2}, x = - 12

Decimal Notation

x = 0.5, x = - 12

Solution steps

(1 - 2 x) (x + 12) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

1 - 2 x = 0 or x + 12 = 0

Solve

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

Solve

x + 12 = 0 : x = - 12

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - 12

- \frac{8}{9 ( 3 ) ^{\frac{1}{3}}}

\frac{3 x + 2}{5} - \frac{x - 2}{35} = 1 - \frac{4 x - 3}{7}

3 x^{2} - x - 44 = 0

2 x^{2} - 9 x - 35 = 0

(2 + 8 i)^{2}

What are the solutions to the equation (1-2x)(x+12)=0 ?
The solutions to the equation (1-2x)(x+12)=0 are x= 1/2 ,x=-12
Find the zeros of (1-2x)(x+12)=0
The zeros of (1-2x)(x+12)=0 are x= 1/2 ,x=-12