What are the solutions to the equation (2x-1)(3x+5)=(x+2)(2x-1) ?

The solutions to the equation (2x-1)(3x+5)=(x+2)(2x-1) are x= 1/2 ,x=-3/2

Find the zeros of (2x-1)(3x+5)=(x+2)(2x-1)

The zeros of (2x-1)(3x+5)=(x+2)(2x-1) are x= 1/2 ,x=-3/2

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(2 x - 1) (3 x + 5) = (x + 2) (2 x - 1)

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

Decimal Notation

x = 0.5, x = - 1.5

Solution steps

(2 x - 1) (3 x + 5) = (x + 2) (2 x - 1)

Move

(x + 2) (2 x - 1)

to the left side

(2 x - 1) (3 x + 5) - (x + 2) (2 x - 1) = 0

Factor

(2 x - 1) (3 x + 5) - (x + 2) (2 x - 1) : (2 x - 1) (2 x + 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 x - 1 = 0 : x = \frac{1}{2}

Solve

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

The solutions to the quadratic equation are:

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

(256 x^{16})^{\frac{1}{4}}

- 11 \cdot - 7

\frac{1}{3} (6)^{3}

\frac{5 x - 9}{3} + \frac{x}{2} = 10

(4 x + 9 i) (x - 1 i)

What are the solutions to the equation (2x-1)(3x+5)=(x+2)(2x-1) ?
The solutions to the equation (2x-1)(3x+5)=(x+2)(2x-1) are x= 1/2 ,x=-3/2
Find the zeros of (2x-1)(3x+5)=(x+2)(2x-1)
The zeros of (2x-1)(3x+5)=(x+2)(2x-1) are x= 1/2 ,x=-3/2