What are the solutions to the equation (2p+3)(p-5)=0 ?

The solutions to the equation (2p+3)(p-5)=0 are p=-3/2 ,p=5

Find the zeros of (2p+3)(p-5)=0

The zeros of (2p+3)(p-5)=0 are p=-3/2 ,p=5

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(2 p + 3) (p - 5) = 0

p = - \frac{3}{2}, p = 5

Decimal Notation

p = - 1.5, p = 5

Solution steps

(2 p + 3) (p - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 p + 3 = 0 or p - 5 = 0

Solve

2 p + 3 = 0 : p = - \frac{3}{2}

Solve

p - 5 = 0 : p = 5

The solutions to the quadratic equation are:

p = - \frac{3}{2}, p = 5

(z + 6)^{2} = 64

45 m^{2} + 0 + 4 = 0

- 4 x^{2} + 30 x + 54 = 0

34 x^{2} + 100 x - 125 = 0

8 π = 2 π + r (r + 3)

What are the solutions to the equation (2p+3)(p-5)=0 ?
The solutions to the equation (2p+3)(p-5)=0 are p=-3/2 ,p=5
Find the zeros of (2p+3)(p-5)=0
The zeros of (2p+3)(p-5)=0 are p=-3/2 ,p=5