What are the solutions to the equation (x-2)(x+4)=0 ?

The solutions to the equation (x-2)(x+4)=0 are x=2,x=-4

Find the zeros of (x-2)(x+4)=0

The zeros of (x-2)(x+4)=0 are x=2,x=-4

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

x = 2, x = - 4

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 4 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 4 = 0 : x = - 4

The solutions to the quadratic equation are:

x = 2, x = - 4

r^{2} + 8 r + 32 = 0

11 x = 8 x - 6

52 = n - \frac{1}{5} n

4^{9}

\frac{1}{2} (11 + 15)

What are the solutions to the equation (x-2)(x+4)=0 ?
The solutions to the equation (x-2)(x+4)=0 are x=2,x=-4
Find the zeros of (x-2)(x+4)=0
The zeros of (x-2)(x+4)=0 are x=2,x=-4