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

(5 x + 2 y)^{2} - (3 x - y)^{2}

- \frac{1}{3} (2) - \frac{3 - 4 ( 2 )}{2}

x^{2} = \frac{7}{36}

b^{3} - 2 = 2

∣ 2 x - 3 ∣ + ∣ x ∣ = 5

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