What are the solutions to the equation (2x-5)(x+4)=0 ?

The solutions to the equation (2x-5)(x+4)=0 are x= 5/2 ,x=-4

Find the zeros of (2x-5)(x+4)=0

The zeros of (2x-5)(x+4)=0 are x= 5/2 ,x=-4

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(2 x - 5) (x + 4) = 0

x = \frac{5}{2}, x = - 4

Decimal Notation

x = 2.5, x = - 4

Solution steps

(2 x - 5) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x - 5 = 0 or x + 4 = 0

Solve

2 x - 5 = 0 : x = \frac{5}{2}

Solve

x + 4 = 0 : x = - 4

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = - 4

0 = x^{2} + 3 x - 5

16 x^{2} + 36 = - 48 x

36 π = 4 π r^{2}

2 x^{2} - 8 x = 0

p, q = 400 - p^{2}

What are the solutions to the equation (2x-5)(x+4)=0 ?
The solutions to the equation (2x-5)(x+4)=0 are x= 5/2 ,x=-4
Find the zeros of (2x-5)(x+4)=0
The zeros of (2x-5)(x+4)=0 are x= 5/2 ,x=-4