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

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

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

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

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

x = - 4, x = 5

Solution steps

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

Move

(x + 4) (x + 2)

to the left side

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

Factor

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

(x + 4) (x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0 or x - 5 = 0

Solve

x + 4 = 0 : x = - 4

Solve

x - 5 = 0 : x = 5

The solutions to the quadratic equation are:

x = - 4, x = 5

9 = \frac{n ( n - 3 )}{2}

4 d^{2} - 19 d + 21 = 2

x^{2} - 7 x + 20 = 0

1 2^{2} + b^{2} = 1 8^{2}

- 4 x^{2} + 2 x + 8 = 0

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