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

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

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

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

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

x = 2, x = - 5

Solution steps

(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 x)^{\frac{4}{5}}

- 12 + \frac{n}{10} > - 7

6 (7 x - 1) = 43 x

(\frac{2}{2})^{2}

\frac{n - 1}{4 n ^{2} - 40 n}

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