What are the solutions to the equation 3(x-2)(x+1)-2(x-1)^2=4 ?

The solutions to the equation 3(x-2)(x+1)-2(x-1)^2=4 are x=3,x=-4

Find the zeros of 3(x-2)(x+1)-2(x-1)^2=4

The zeros of 3(x-2)(x+1)-2(x-1)^2=4 are x=3,x=-4

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3 (x - 2) (x + 1) - 2 (x - 1)^{2} = 4

x = 3, x = - 4

Solution steps

3 (x - 2) (x + 1) - 2 (x - 1)^{2} = 4

Expand

3 (x - 2) (x + 1) - 2 (x - 1)^{2} : x^{2} + x - 8

x^{2} + x - 8 = 4

Move

4

to the left side

x^{2} + x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 1}, x_{2} = \frac{- 1 - 7}{2 \cdot 1}

x = \frac{- 1 + 7}{2 \cdot 1} : 3

x = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 3, x = - 4

64 x^{2} + 144 x + 81 = 25

(n - 2)^{2} = 5

- 2 x^{2} + 7 x = 6

9 x^{2} + 12 x + 7 = 0

6 s^{2} + 18 s = 0

What are the solutions to the equation 3(x-2)(x+1)-2(x-1)^2=4 ?
The solutions to the equation 3(x-2)(x+1)-2(x-1)^2=4 are x=3,x=-4
Find the zeros of 3(x-2)(x+1)-2(x-1)^2=4
The zeros of 3(x-2)(x+1)-2(x-1)^2=4 are x=3,x=-4