What are the solutions to the equation 10=x^2-4x+1 ?

The solutions to the equation 10=x^2-4x+1 are x=2+sqrt(13),x=2-sqrt(13)

Find the zeros of 10=x^2-4x+1

The zeros of 10=x^2-4x+1 are x=2+sqrt(13),x=2-sqrt(13)

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10 = x^{2} - 4 x + 1

x = 2 + 13, x = 2 - 13

Decimal Notation

x = 5.60555 \dots, x = - 1.60555 \dots

Solution steps

10 = x^{2} - 4 x + 1

Switch sides

x^{2} - 4 x + 1 = 10

Move

10

to the left side

x^{2} - 4 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 9) = 213

x_{1, 2} = \frac{- ( - 4 ) \pm 2 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 13}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 13}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 13}{2 \cdot 1} : 2 + 13

x = \frac{- ( - 4 ) - 2 13}{2 \cdot 1} : 2 - 13

The solutions to the quadratic equation are:

x = 2 + 13, x = 2 - 13

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

x^{2} - 9 x = 4

5 y^{2} + 9 y - 2 = 0

x^{2} = 6 x + 4

x^{2} - 18 x - 11 = 0

What are the solutions to the equation 10=x^2-4x+1 ?
The solutions to the equation 10=x^2-4x+1 are x=2+sqrt(13),x=2-sqrt(13)
Find the zeros of 10=x^2-4x+1
The zeros of 10=x^2-4x+1 are x=2+sqrt(13),x=2-sqrt(13)