What are the solutions to the equation x^2-x+1=7 ?

The solutions to the equation x^2-x+1=7 are x=3,x=-2

Find the zeros of x^2-x+1=7

The zeros of x^2-x+1=7 are x=3,x=-2

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x^{2} - x + 1 = 7

x = 3, x = - 2

Solution steps

x^{2} - x + 1 = 7

Move

7

to the left side

x^{2} - x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 6) = 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} : 3

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

The solutions to the quadratic equation are:

x = 3, x = - 2

x, z = 16 - x^{2} - y^{2}

- 2 x^{2} + 2 x - 1 = 0

- 36 x^{2} + 100 = 0

144 x^{2} - 100 = 0

s^{2} + 8 s + 25

What are the solutions to the equation x^2-x+1=7 ?
The solutions to the equation x^2-x+1=7 are x=3,x=-2
Find the zeros of x^2-x+1=7
The zeros of x^2-x+1=7 are x=3,x=-2