What are the solutions to the equation 24(x+4)(x+3)-24(x+2)(x+5)=(x+5)(x+3) ?

The solutions to the equation 24(x+4)(x+3)-24(x+2)(x+5)=(x+5)(x+3) are x=3,x=-11

Find the zeros of 24(x+4)(x+3)-24(x+2)(x+5)=(x+5)(x+3)

The zeros of 24(x+4)(x+3)-24(x+2)(x+5)=(x+5)(x+3) are x=3,x=-11

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24 (x + 4) (x + 3) - 24 (x + 2) (x + 5) = (x + 5) (x + 3)

x = 3, x = - 11

Solution steps

24 (x + 4) (x + 3) - 24 (x + 2) (x + 5) = (x + 5) (x + 3)

Expand

24 (x + 4) (x + 3) - 24 (x + 2) (x + 5) : 48

Expand

(x + 5) (x + 3) : x^{2} + 8 x + 15

48 = x^{2} + 8 x + 15

Switch sides

x^{2} + 8 x + 15 = 48

Move

48

to the left side

x^{2} + 8 x - 33 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 33 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 33) = 14

x_{1, 2} = \frac{- 8 \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 14}{2 \cdot 1}, x_{2} = \frac{- 8 - 14}{2 \cdot 1}

x = \frac{- 8 + 14}{2 \cdot 1} : 3

x = \frac{- 8 - 14}{2 \cdot 1} : - 11

The solutions to the quadratic equation are:

x = 3, x = - 11

x^{2} - 18 + 56 = 0

x^{2} + x - 1 = 0

0 = 3 t^{2} - 12 t + 9

3 x^{2} - 4 x = 9

(2 x + 1)^{2} = 32

What are the solutions to the equation 24(x+4)(x+3)-24(x+2)(x+5)=(x+5)(x+3) ?
The solutions to the equation 24(x+4)(x+3)-24(x+2)(x+5)=(x+5)(x+3) are x=3,x=-11
Find the zeros of 24(x+4)(x+3)-24(x+2)(x+5)=(x+5)(x+3)
The zeros of 24(x+4)(x+3)-24(x+2)(x+5)=(x+5)(x+3) are x=3,x=-11