What are the solutions to the equation (r-3)(r+2)=0 ?

The solutions to the equation (r-3)(r+2)=0 are r=3,r=-2

Find the zeros of (r-3)(r+2)=0

The zeros of (r-3)(r+2)=0 are r=3,r=-2

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(r - 3) (r + 2) = 0

r = 3, r = - 2

Solution steps

(r - 3) (r + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r - 3 = 0 or r + 2 = 0

Solve

r - 3 = 0 : r = 3

Solve

r + 2 = 0 : r = - 2

The solutions to the quadratic equation are:

r = 3, r = - 2

x^{2} - 3 x - 40 \leq 0

2 7^{\frac{- 4}{3}}

48 t u - 90 t + 32 u - 60

6 x^{3} - 4 x^{2} - 2 x

5 x^{2} - 320

What are the solutions to the equation (r-3)(r+2)=0 ?
The solutions to the equation (r-3)(r+2)=0 are r=3,r=-2
Find the zeros of (r-3)(r+2)=0
The zeros of (r-3)(r+2)=0 are r=3,r=-2