What are the solutions to the equation (3x-2)(4x+8)=10(x-2)(x+2) ?

The solutions to the equation (3x-2)(4x+8)=10(x-2)(x+2) are x=-2,x=-6

Find the zeros of (3x-2)(4x+8)=10(x-2)(x+2)

The zeros of (3x-2)(4x+8)=10(x-2)(x+2) are x=-2,x=-6

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(3 x - 2) (4 x + 8) = 10 (x - 2) (x + 2)

x = - 2, x = - 6

Solution steps

(3 x - 2) (4 x + 8) = 10 (x - 2) (x + 2)

Move

10 (x - 2) (x + 2)

to the left side

(3 x - 2) (4 x + 8) - 10 (x - 2) (x + 2) = 0

Factor

(3 x - 2) (4 x + 8) - 10 (x - 2) (x + 2) : 2 (x + 2) (x + 6)

2 (x + 2) (x + 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x + 6 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x + 6 = 0 : x = - 6

The solutions to the quadratic equation are:

x = - 2, x = - 6

4 h^{2} - 20 h + 25 = 0

x, x^{2} y^{2} + x y = 6

- 3 x^{2} + 6 = 18

x^{2} + 6 0^{2} = 6 1^{2}

4 x^{2} - 20 x + 25 = 0

What are the solutions to the equation (3x-2)(4x+8)=10(x-2)(x+2) ?
The solutions to the equation (3x-2)(4x+8)=10(x-2)(x+2) are x=-2,x=-6
Find the zeros of (3x-2)(4x+8)=10(x-2)(x+2)
The zeros of (3x-2)(4x+8)=10(x-2)(x+2) are x=-2,x=-6