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

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

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

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

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

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

Decimal Notation

x = 1.25, x = 2

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

4 x - 5 = 0 : x = \frac{5}{4}

Solve

- x + 2 = 0 : x = 2

The solutions to the quadratic equation are:

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

(5 x + 7)^{2} - (- 2 x + 5)^{2} = 0

x^{2} + 20 x + c

y, 4 x^{2} - 16 x + 4 y^{2} + 16 = 0

x^{2} - 5 x + \frac{4}{9} = 0

2 (x + 4)^{2} + 15 = 17

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