What are the solutions to the equation (-2x+1)(-2x-7)=0 ?

The solutions to the equation (-2x+1)(-2x-7)=0 are x= 1/2 ,x=-7/2

Find the zeros of (-2x+1)(-2x-7)=0

The zeros of (-2x+1)(-2x-7)=0 are x= 1/2 ,x=-7/2

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(- 2 x + 1) (- 2 x - 7) = 0

x = \frac{1}{2}, x = - \frac{7}{2}

Decimal Notation

x = 0.5, x = - 3.5

Solution steps

(- 2 x + 1) (- 2 x - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

- 2 x + 1 = 0 or - 2 x - 7 = 0

Solve

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

Solve

- 2 x - 7 = 0 : x = - \frac{7}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - \frac{7}{2}

w^{2} + 7 w + 4 = 0

z^{2} + 0.35 z - 1.125 = 0

x^{2} - 7 x - 1 = - 1

x^{2} + 6 x + 8

(x + 4) (2 x - a) = 0

What are the solutions to the equation (-2x+1)(-2x-7)=0 ?
The solutions to the equation (-2x+1)(-2x-7)=0 are x= 1/2 ,x=-7/2
Find the zeros of (-2x+1)(-2x-7)=0
The zeros of (-2x+1)(-2x-7)=0 are x= 1/2 ,x=-7/2