What are the solutions to the equation x^2-6(x-1)=7(1-x) ?

The solutions to the equation x^2-6(x-1)=7(1-x) are x=(-1+sqrt(5))/2 ,x=(-1-sqrt(5))/2

Find the zeros of x^2-6(x-1)=7(1-x)

The zeros of x^2-6(x-1)=7(1-x) are x=(-1+sqrt(5))/2 ,x=(-1-sqrt(5))/2

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x^{2} - 6 (x - 1) = 7 (1 - x)

x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

Solution steps

x^{2} - 6 (x - 1) = 7 (1 - x)

Expand

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

Expand

7 (1 - x) : 7 - 7 x

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

Move

7 x

to the left side

x^{2} + x + 6 = 7

Move

7

to the left side

x^{2} + x - 1 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

x_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 5}{2 \cdot 1}, x_{2} = \frac{- 1 - 5}{2 \cdot 1}

x = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

x = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

x^{2} - x = - 3 x + 3

x, x (x - c) = a

4 x^{2} + 13 x = 12

2 (x + 3)^{2} + 4 = 0

(x - 11)^{2} = 0

What are the solutions to the equation x^2-6(x-1)=7(1-x) ?
The solutions to the equation x^2-6(x-1)=7(1-x) are x=(-1+sqrt(5))/2 ,x=(-1-sqrt(5))/2
Find the zeros of x^2-6(x-1)=7(1-x)
The zeros of x^2-6(x-1)=7(1-x) are x=(-1+sqrt(5))/2 ,x=(-1-sqrt(5))/2