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

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

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

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

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

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

Decimal Notation

k = - 1.5, k = 5

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

k - 5 = 0 : k = 5

The solutions to the quadratic equation are:

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

(3 x^{3} y^{2}) (- 6 y^{5})

\frac{2 π}{60}

3 (t + 1) = - \frac{1}{2} - 5 t

x^{2} - 8 x + 12 < 0

19 x^{3} y^{2} + 57 x^{5} y

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