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

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

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

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

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

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

Decimal Notation

x = - 0.5, x = 7

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

x - 7 = 0 : x = 7

The solutions to the quadratic equation are:

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

x, x^{2} + m x + m + 5 = 0

λ^{2} - 4 λ + 3 = 0

x, x^{2} y - 4 y + 4 = 0

(x - 3)^{3} = x^{3} + 5

b, V = \frac{1}{3} b^{2} h

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