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

1 + 3

36 p^{2} + 14 p

k - 6 < 3

\frac{18 a ^{3} b ^{2}}{2 ab}

lo g_{\frac{1}{27}} (81) = \frac{2}{3} x + 9

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