What are the solutions to the equation (5x+1)(x-2)=0 ?

The solutions to the equation (5x+1)(x-2)=0 are x=-1/5 ,x=2

Find the zeros of (5x+1)(x-2)=0

The zeros of (5x+1)(x-2)=0 are x=-1/5 ,x=2

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(5 x + 1) (x - 2) = 0

x = - \frac{1}{5}, x = 2

Decimal Notation

x = - 0.2, x = 2

Solution steps

(5 x + 1) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

5 x + 1 = 0 or x - 2 = 0

Solve

5 x + 1 = 0 : x = - \frac{1}{5}

Solve

x - 2 = 0 : x = 2

The solutions to the quadratic equation are:

x = - \frac{1}{5}, x = 2

4 m^{2} - m - 3 = - m^{2}

(x + 5) (x + 6)

196 x^{2} + 49 x + 2 = 0

x, x^{2} + k x + 36 = 0

(x + 1)^{2} = - 48

What are the solutions to the equation (5x+1)(x-2)=0 ?
The solutions to the equation (5x+1)(x-2)=0 are x=-1/5 ,x=2
Find the zeros of (5x+1)(x-2)=0
The zeros of (5x+1)(x-2)=0 are x=-1/5 ,x=2