What are the solutions to the equation solvefor p,x^2-(2p-1)x+p^2-p-6=0 ?

The solutions to the equation solvefor p,x^2-(2p-1)x+p^2-p-6=0 are p=x+3,p=x-2

Find the zeros of solvefor p,x^2-(2p-1)x+p^2-p-6=0

The zeros of solvefor p,x^2-(2p-1)x+p^2-p-6=0 are p=x+3,p=x-2

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solve for

p, x^{2} - (2 p - 1) x + p^{2} - p - 6 = 0

p = x + 3, p = x - 2

Solution steps

x^{2} - (2 p - 1) x + p^{2} - p - 6 = 0

Expand

x^{2} - (2 p - 1) x + p^{2} - p - 6 : x^{2} - 2 x p + x + p^{2} - p - 6

x^{2} - 2 x p + x + p^{2} - p - 6 = 0

Write in the standard form

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

p^{2} - (2 x + 1) p + x^{2} + x - 6 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 2 x - 1 ) \pm ( - 2 x - 1 ) ^{2} - 4 \cdot 1 \cdot ( x ^{2} + x - 6 )}{2 \cdot 1}

(- 2 x - 1)^{2} - 4 \cdot 1 \cdot (x^{2} + x - 6) = 5

p_{1, 2} = \frac{- ( - 2 x - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- ( - 2 x - 1 ) + 5}{2 \cdot 1}, p_{2} = \frac{- ( - 2 x - 1 ) - 5}{2 \cdot 1}

p = \frac{- ( - 2 x - 1 ) + 5}{2 \cdot 1} : x + 3

p = \frac{- ( - 2 x - 1 ) - 5}{2 \cdot 1} : x - 2

The solutions to the quadratic equation are:

p = x + 3, p = x - 2

x^{2} - 1.4 x + 0.45 = 0

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

x^{2} - 12 x - 5 = 0

x^{2} - 12 x - 5 = 7

36 = \frac{x ^{2} - x}{2}

What are the solutions to the equation solvefor p,x^2-(2p-1)x+p^2-p-6=0 ?
The solutions to the equation solvefor p,x^2-(2p-1)x+p^2-p-6=0 are p=x+3,p=x-2
Find the zeros of solvefor p,x^2-(2p-1)x+p^2-p-6=0
The zeros of solvefor p,x^2-(2p-1)x+p^2-p-6=0 are p=x+3,p=x-2