What are the solutions to the equation (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0 ?

The solutions to the equation (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0 are x=6,x=0

Find the zeros of (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0

The zeros of (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0 are x=6,x=0

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(8 - 5 x) (x + 2) + 4 (x - 2) (x + 1) + 2 (x - 2) (x + 2) = 0

x = 6, x = 0

Solution steps

(8 - 5 x) (x + 2) + 4 (x - 2) (x + 1) + 2 (x - 2) (x + 2) = 0

Expand

(8 - 5 x) (x + 2) + 4 (x - 2) (x + 1) + 2 (x - 2) (x + 2) : x^{2} - 6 x

x^{2} - 6 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 0 = 6

x_{1, 2} = \frac{- ( - 6 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 6}{2 \cdot 1}

x = \frac{- ( - 6 ) + 6}{2 \cdot 1} : 6

x = \frac{- ( - 6 ) - 6}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 6, x = 0

\frac{5 a ^{2}}{5 a}

3 (x - 4) < 2 (2 x - 1)

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

What are the solutions to the equation (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0 ?
The solutions to the equation (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0 are x=6,x=0
Find the zeros of (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0
The zeros of (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0 are x=6,x=0