What are the solutions to the equation (2x+3)(2x-3)=4(2x+3) ?

The solutions to the equation (2x+3)(2x-3)=4(2x+3) are x=-3/2 ,x= 7/2

Find the zeros of (2x+3)(2x-3)=4(2x+3)

The zeros of (2x+3)(2x-3)=4(2x+3) are x=-3/2 ,x= 7/2

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(2 x + 3) (2 x - 3) = 4 (2 x + 3)

x = - \frac{3}{2}, x = \frac{7}{2}

Decimal Notation

x = - 1.5, x = 3.5

Solution steps

(2 x + 3) (2 x - 3) = 4 (2 x + 3)

Move

4 (2 x + 3)

to the left side

(2 x + 3) (2 x - 3) - 4 (2 x + 3) = 0

Factor

(2 x + 3) (2 x - 3) - 4 (2 x + 3) : (2 x + 3) (2 x - 7)

(2 x + 3) (2 x - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x + 3 = 0 or 2 x - 7 = 0

Solve

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

Solve

2 x - 7 = 0 : x = \frac{7}{2}

The solutions to the quadratic equation are:

x = - \frac{3}{2}, x = \frac{7}{2}

3 + 5 x = - 9 x^{2}

4 x^{2} + 8 x + 1 = 2 x^{2}

x^{2} - 2 x + 5 = 85

- 3 x^{2} - 24 x - 46

65 x^{2} - 1 = x^{2}

What are the solutions to the equation (2x+3)(2x-3)=4(2x+3) ?
The solutions to the equation (2x+3)(2x-3)=4(2x+3) are x=-3/2 ,x= 7/2
Find the zeros of (2x+3)(2x-3)=4(2x+3)
The zeros of (2x+3)(2x-3)=4(2x+3) are x=-3/2 ,x= 7/2