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

The solutions to the equation x^2-2(m-2)x+(m^2-2m-3)=0 are x=m-2+sqrt(-2m+7),x=m-2-sqrt(-2m+7)

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

The zeros of x^2-2(m-2)x+(m^2-2m-3)=0 are x=m-2+sqrt(-2m+7),x=m-2-sqrt(-2m+7)

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

x = m - 2 + - 2 m + 7, x = m - 2 - - 2 m + 7

Solution steps

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

Expand

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 2 m + 4)^{2} - 4 \cdot 1 \cdot (m^{2} - 2 m - 3) : 2 - 2 m + 7

x_{1, 2} = \frac{- ( - 2 m + 4 ) \pm 2 - 2 m + 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 m + 4 ) + 2 - 2 m + 7}{2 \cdot 1}, x_{2} = \frac{- ( - 2 m + 4 ) - 2 - 2 m + 7}{2 \cdot 1}

x = \frac{- ( - 2 m + 4 ) + 2 - 2 m + 7}{2 \cdot 1} : m - 2 + - 2 m + 7

x = \frac{- ( - 2 m + 4 ) - 2 - 2 m + 7}{2 \cdot 1} : m - 2 - - 2 m + 7

The solutions to the quadratic equation are:

x = m - 2 + - 2 m + 7, x = m - 2 - - 2 m + 7

x^{2} + 16 x - 225 = 0

x, (p - 1) x^{2} + 2 p x + p - 1 = 0

m = x^{2} a + x^{2} b

2 x^{2} + 6 = - 1, 3 x = 0

5 r^{2} - 80 = 0

What are the solutions to the equation x^2-2(m-2)x+(m^2-2m-3)=0 ?
The solutions to the equation x^2-2(m-2)x+(m^2-2m-3)=0 are x=m-2+sqrt(-2m+7),x=m-2-sqrt(-2m+7)
Find the zeros of x^2-2(m-2)x+(m^2-2m-3)=0
The zeros of x^2-2(m-2)x+(m^2-2m-3)=0 are x=m-2+sqrt(-2m+7),x=m-2-sqrt(-2m+7)