What are the solutions to the equation (2x-3)(x+1)=0 ?

The solutions to the equation (2x-3)(x+1)=0 are x= 3/2 ,x=-1

Find the zeros of (2x-3)(x+1)=0

The zeros of (2x-3)(x+1)=0 are x= 3/2 ,x=-1

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(2 x - 3) (x + 1) = 0

x = \frac{3}{2}, x = - 1

Solution steps

(2 x - 3) (x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x - 3 = 0 or x + 1 = 0

Solve

2 x - 3 = 0 : x = \frac{3}{2}

Solve

x + 1 = 0 : x = - 1

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = - 1

x^{2} + 5 x = - 2

\frac{3}{5} \cdot 8

\frac{\frac{3}{2}}{2}

lo g_{5} (625) = y

b^{6} + 1

What are the solutions to the equation (2x-3)(x+1)=0 ?
The solutions to the equation (2x-3)(x+1)=0 are x= 3/2 ,x=-1
Find the zeros of (2x-3)(x+1)=0
The zeros of (2x-3)(x+1)=0 are x= 3/2 ,x=-1