What are the solutions to the equation 8=x^2-2x+7 ?

The solutions to the equation 8=x^2-2x+7 are x=1+sqrt(2),x=1-sqrt(2)

Find the zeros of 8=x^2-2x+7

The zeros of 8=x^2-2x+7 are x=1+sqrt(2),x=1-sqrt(2)

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

x = 1 + 2, x = 1 - 2

Decimal Notation

x = 2.41421 \dots, x = - 0.41421 \dots

Solution steps

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

Switch sides

x^{2} - 2 x + 7 = 8

Move

8

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 2, x = 1 - 2

(- \frac{4}{5})^{3}

12 ∣ x ∣ - 1 = 3

(2 x^{2} + x) \geq 1

8 n \leq - 64

(x^{2} + x + 4)^{2} + 8 x (x^{2} + x + 4) + 15 x^{2}

What are the solutions to the equation 8=x^2-2x+7 ?
The solutions to the equation 8=x^2-2x+7 are x=1+sqrt(2),x=1-sqrt(2)
Find the zeros of 8=x^2-2x+7
The zeros of 8=x^2-2x+7 are x=1+sqrt(2),x=1-sqrt(2)