What are the solutions to the equation 120+(x+2)(x+4)=24(x+4) ?

The solutions to the equation 120+(x+2)(x+4)=24(x+4) are x=16,x=2

Find the zeros of 120+(x+2)(x+4)=24(x+4)

The zeros of 120+(x+2)(x+4)=24(x+4) are x=16,x=2

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120 + (x + 2) (x + 4) = 24 (x + 4)

x = 16, x = 2

Solution steps

120 + (x + 2) (x + 4) = 24 (x + 4)

Expand

120 + (x + 2) (x + 4) : x^{2} + 6 x + 128

Expand

24 (x + 4) : 24 x + 96

x^{2} + 6 x + 128 = 24 x + 96

Move

96

to the left side

x^{2} + 6 x + 32 = 24 x

Move

24 x

to the left side

x^{2} - 18 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 32 = 14

x_{1, 2} = \frac{- ( - 18 ) \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 14}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 14}{2 \cdot 1}

x = \frac{- ( - 18 ) + 14}{2 \cdot 1} : 16

x = \frac{- ( - 18 ) - 14}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 16, x = 2

s^{2} + 9 s + 17 = 0

(4)^{3} - k^{2} (4)^{2} - 8 k (4) - 16 = 0

8 a^{2} - 10 a + 2 = 0

16 a^{2} - 25 = 0

x, p x^{2} - p (p + 3) x + 2 p (p + 1) = 0

What are the solutions to the equation 120+(x+2)(x+4)=24(x+4) ?
The solutions to the equation 120+(x+2)(x+4)=24(x+4) are x=16,x=2
Find the zeros of 120+(x+2)(x+4)=24(x+4)
The zeros of 120+(x+2)(x+4)=24(x+4) are x=16,x=2