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

The solutions to the equation x=(7-x)(6-x) are x=7+sqrt(7),x=7-sqrt(7)

Find the zeros of x=(7-x)(6-x)

The zeros of x=(7-x)(6-x) are x=7+sqrt(7),x=7-sqrt(7)

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x = (7 - x) (6 - x)

x = 7 + 7, x = 7 - 7

Decimal Notation

x = 9.64575 \dots, x = 4.35424 \dots

Solution steps

x = (7 - x) (6 - x)

Expand

(7 - x) (6 - x) : 42 - 13 x + x^{2}

x = 42 - 13 x + x^{2}

Switch sides

42 - 13 x + x^{2} = x

Move

x

to the left side

x^{2} - 14 x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 42 = 27

x_{1, 2} = \frac{- ( - 14 ) \pm 2 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2 7}{2 \cdot 1}, x_{2} = \frac{- ( - 14 ) - 2 7}{2 \cdot 1}

x = \frac{- ( - 14 ) + 2 7}{2 \cdot 1} : 7 + 7

x = \frac{- ( - 14 ) - 2 7}{2 \cdot 1} : 7 - 7

The solutions to the quadratic equation are:

x = 7 + 7, x = 7 - 7

x^{2} - \frac{1}{4} x - \frac{35}{64} = 0

v^{2} + v - 19 = 4 v

- 2 x^{2} + 1 = 5 x

3 q^{2} + 10 q + 1 = q

3 m^{2} + 20 m + 25 = 0

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