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

The solutions to the equation (x+3)(2x-1)=(2x-1)(3x-5) are x=4,x= 1/2

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

The zeros of (x+3)(2x-1)=(2x-1)(3x-5) are x=4,x= 1/2

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

x = 4, x = \frac{1}{2}

Decimal Notation

x = 4, x = 0.5

Solution steps

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

Move

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

to the left side

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

Factor

(x + 3) (2 x - 1) - (2 x - 1) (3 x - 5) : - 2 (x - 4) (2 x - 1)

- 2 (x - 4) (2 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 4 = 0 or 2 x - 1 = 0

Solve

x - 4 = 0 : x = 4

Solve

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

The solutions to the quadratic equation are:

x = 4, x = \frac{1}{2}

1 x^{2} + 6400 x - 115200000 = 0

0 = 150 + 90 t + 4.9 t^{2}

θ^{2} = 4 0^{\circ}

x, y^{2} - x^{2} = k^{2}

w, r^{2} + w^{2} = 0

What are the solutions to the equation (x+3)(2x-1)=(2x-1)(3x-5) ?
The solutions to the equation (x+3)(2x-1)=(2x-1)(3x-5) are x=4,x= 1/2
Find the zeros of (x+3)(2x-1)=(2x-1)(3x-5)
The zeros of (x+3)(2x-1)=(2x-1)(3x-5) are x=4,x= 1/2