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

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

Find the zeros of (n+2)(2n+5)=0

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

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(n + 2) (2 n + 5) = 0

n = - 2, n = - \frac{5}{2}

Solution steps

(n + 2) (2 n + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n + 2 = 0 or 2 n + 5 = 0

Solve

n + 2 = 0 : n = - 2

Solve

2 n + 5 = 0 : n = - \frac{5}{2}

The solutions to the quadratic equation are:

n = - 2, n = - \frac{5}{2}

(9^{\frac{2}{3}}) (3) (3^{\frac{2}{3}})

6 - 3 x = - 3

64 x^{2} - 25

- 8 c + 3 = c + 8 - 10 c

16 x^{3} - 12 x^{2} + 20 x - 15

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