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

The solutions to the equation x^2-x+1=9 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2

Find the zeros of x^2-x+1=9

The zeros of x^2-x+1=9 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2

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x^{2} - x + 1 = 9

x = \frac{1 + 33}{2}, x = \frac{1 - 33}{2}

Decimal Notation

x = 3.37228 \dots, x = - 2.37228 \dots

Solution steps

x^{2} - x + 1 = 9

Move

9

to the left side

x^{2} - x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 8) = 33

x_{1, 2} = \frac{- ( - 1 ) \pm 33}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 33}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 33}{2 \cdot 1}

x = \frac{- ( - 1 ) + 33}{2 \cdot 1} : \frac{1 + 33}{2}

x = \frac{- ( - 1 ) - 33}{2 \cdot 1} : \frac{1 - 33}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 33}{2}, x = \frac{1 - 33}{2}

7 + 4 ∣ 5 x - 8 ∣ = 15

c = 3 0^{2} + 1 8^{2}

(\frac{8}{6})^{2}

\frac{2}{4} \cdot \frac{1}{2}

r^{2} + 12 r + 52 = 0

What are the solutions to the equation x^2-x+1=9 ?
The solutions to the equation x^2-x+1=9 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2
Find the zeros of x^2-x+1=9
The zeros of x^2-x+1=9 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2