What are the solutions to the equation (x-2)^2+(1-3x)^2=5 ?

The solutions to the equation (x-2)^2+(1-3x)^2=5 are x=1,x=0

Find the zeros of (x-2)^2+(1-3x)^2=5

The zeros of (x-2)^2+(1-3x)^2=5 are x=1,x=0

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(x - 2)^{2} + (1 - 3 x)^{2} = 5

x = 1, x = 0

Solution steps

(x - 2)^{2} + (1 - 3 x)^{2} = 5

Expand

(x - 2)^{2} + (1 - 3 x)^{2} : 10 x^{2} - 10 x + 5

10 x^{2} - 10 x + 5 = 5

Move

5

to the left side

10 x^{2} - 10 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

(- 10)^{2} - 4 \cdot 10 \cdot 0 = 10

x_{1, 2} = \frac{- ( - 10 ) \pm 10}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 10}{2 \cdot 10}, x_{2} = \frac{- ( - 10 ) - 10}{2 \cdot 10}

x = \frac{- ( - 10 ) + 10}{2 \cdot 10} : 1

x = \frac{- ( - 10 ) - 10}{2 \cdot 10} : 0

The solutions to the quadratic equation are:

x = 1, x = 0

x^{2} + 12 x + 36 = 121

(y - 7)^{2} - 6 (y - 7) + 5 = 0

59 = y^{2} - 10 y

p^{2} - 5 p - 24 = 0

8 x = 9 - x^{2}

What are the solutions to the equation (x-2)^2+(1-3x)^2=5 ?
The solutions to the equation (x-2)^2+(1-3x)^2=5 are x=1,x=0
Find the zeros of (x-2)^2+(1-3x)^2=5
The zeros of (x-2)^2+(1-3x)^2=5 are x=1,x=0