What are the solutions to the equation (x+9)(x-2)=1 ?

The solutions to the equation (x+9)(x-2)=1 are x=(-7+5sqrt(5))/2 ,x=(-7-5sqrt(5))/2

Find the zeros of (x+9)(x-2)=1

The zeros of (x+9)(x-2)=1 are x=(-7+5sqrt(5))/2 ,x=(-7-5sqrt(5))/2

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(x + 9) (x - 2) = 1

x = \frac{- 7 + 5 5}{2}, x = \frac{- 7 - 5 5}{2}

Decimal Notation

x = 2.09016 \dots, x = - 9.09016 \dots

Solution steps

(x + 9) (x - 2) = 1

Expand

(x + 9) (x - 2) : x^{2} + 7 x - 18

x^{2} + 7 x - 18 = 1

Move

1

to the left side

x^{2} + 7 x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 19 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 19) = 55

x_{1, 2} = \frac{- 7 \pm 5 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 5 5}{2 \cdot 1}, x_{2} = \frac{- 7 - 5 5}{2 \cdot 1}

x = \frac{- 7 + 5 5}{2 \cdot 1} : \frac{- 7 + 5 5}{2}

x = \frac{- 7 - 5 5}{2 \cdot 1} : \frac{- 7 - 5 5}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 5 5}{2}, x = \frac{- 7 - 5 5}{2}

12 a^{3} + 10 a^{2}

x^{2} + 2 x + 17 = 6

x^{2} + 4 x < 12

8 x - 15 = 0

\frac{y ^{- 1}}{y ^{- 2}}

What are the solutions to the equation (x+9)(x-2)=1 ?
The solutions to the equation (x+9)(x-2)=1 are x=(-7+5sqrt(5))/2 ,x=(-7-5sqrt(5))/2
Find the zeros of (x+9)(x-2)=1
The zeros of (x+9)(x-2)=1 are x=(-7+5sqrt(5))/2 ,x=(-7-5sqrt(5))/2