What are the solutions to the equation (2x-5)(x+7)=0 ?

The solutions to the equation (2x-5)(x+7)=0 are x= 5/2 ,x=-7

Find the zeros of (2x-5)(x+7)=0

The zeros of (2x-5)(x+7)=0 are x= 5/2 ,x=-7

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(2 x - 5) (x + 7) = 0

x = \frac{5}{2}, x = - 7

Decimal Notation

x = 2.5, x = - 7

Solution steps

(2 x - 5) (x + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x - 5 = 0 or x + 7 = 0

Solve

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

Solve

x + 7 = 0 : x = - 7

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = - 7

x^{2} + 1 = 3

x, x^{2} y + x^{2} - x y^{2} + x y + 2 x = 0

x^{2} - 6 x + 15 = 6

x^{2} - 14 x + 40 = 0

- 2 x^{2} + 12 x - 16 = 0

What are the solutions to the equation (2x-5)(x+7)=0 ?
The solutions to the equation (2x-5)(x+7)=0 are x= 5/2 ,x=-7
Find the zeros of (2x-5)(x+7)=0
The zeros of (2x-5)(x+7)=0 are x= 5/2 ,x=-7