What are the solutions to the equation 0=(x-2)(x+5) ?

The solutions to the equation 0=(x-2)(x+5) are x=2,x=-5

Find the zeros of 0=(x-2)(x+5)

The zeros of 0=(x-2)(x+5) are x=2,x=-5

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0 = (x - 2) (x + 5)

x = 2, x = - 5

Solution steps

0 = (x - 2) (x + 5)

Switch sides

(x - 2) (x + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 5 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 5 = 0 : x = - 5

The solutions to the quadratic equation are:

x = 2, x = - 5

3 (2 x - 3) = 4 x

\frac{- 36}{- 6}

24 b^{3} \cdot 40 b^{2} \cdot b^{2}

\frac{1 + 4 p}{9} = - 3

\frac{2 3}{15}

What are the solutions to the equation 0=(x-2)(x+5) ?
The solutions to the equation 0=(x-2)(x+5) are x=2,x=-5
Find the zeros of 0=(x-2)(x+5)
The zeros of 0=(x-2)(x+5) are x=2,x=-5