What are the solutions to the equation x^2-(2m-4)x+3-4m+m^2=0 ?

The solutions to the equation x^2-(2m-4)x+3-4m+m^2=0 are x=m-1,x=m-3

Find the zeros of x^2-(2m-4)x+3-4m+m^2=0

The zeros of x^2-(2m-4)x+3-4m+m^2=0 are x=m-1,x=m-3

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x^{2} - (2 m - 4) x + 3 - 4 m + m^{2} = 0

x = m - 1, x = m - 3

Solution steps

x^{2} - (2 m - 4) x + 3 - 4 m + m^{2} = 0

Expand

x^{2} - (2 m - 4) x + 3 - 4 m + m^{2} : x^{2} - 2 m x + 4 x + 3 - 4 m + m^{2}

x^{2} - 2 m x + 4 x + 3 - 4 m + m^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + (- 2 m + 4) x + 3 - 4 m + m^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 m + 4 ) \pm ( - 2 m + 4 ) ^{2} - 4 \cdot 1 \cdot ( 3 - 4 m + m ^{2} )}{2 \cdot 1}

(- 2 m + 4)^{2} - 4 \cdot 1 \cdot (3 - 4 m + m^{2}) = 2

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

Separate the solutions

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

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

x = \frac{- ( - 2 m + 4 ) - 2}{2 \cdot 1} : m - 3

The solutions to the quadratic equation are:

x = m - 1, x = m - 3

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

- x - 3 = - x^{2} - x + 6

3 t^{2} - \frac{1}{2} t = 0

a^{2} - 12 = - 4 a

3 = 51 - 2 x^{2}

What are the solutions to the equation x^2-(2m-4)x+3-4m+m^2=0 ?
The solutions to the equation x^2-(2m-4)x+3-4m+m^2=0 are x=m-1,x=m-3
Find the zeros of x^2-(2m-4)x+3-4m+m^2=0
The zeros of x^2-(2m-4)x+3-4m+m^2=0 are x=m-1,x=m-3