What are the solutions to the equation solvefor x,x^2-(2p-1)x+(p^2-p-20)=0 ?

The solutions to the equation solvefor x,x^2-(2p-1)x+(p^2-p-20)=0 are x=p+4,x=p-5

Find the zeros of solvefor x,x^2-(2p-1)x+(p^2-p-20)=0

The zeros of solvefor x,x^2-(2p-1)x+(p^2-p-20)=0 are x=p+4,x=p-5

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solve for

x, x^{2} - (2 p - 1) x + (p^{2} - p - 20) = 0

x = p + 4, x = p - 5

Solution steps

x^{2} - (2 p - 1) x + (p^{2} - p - 20) = 0

Expand

x^{2} - (2 p - 1) x + (p^{2} - p - 20) : x^{2} - 2 x p + x + p^{2} - p - 20

x^{2} - 2 x p + x + p^{2} - p - 20 = 0

Write in the standard form

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

x^{2} + (- 2 p + 1) x + p^{2} - p - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 p + 1 ) \pm ( - 2 p + 1 ) ^{2} - 4 \cdot 1 \cdot ( p ^{2} - p - 20 )}{2 \cdot 1}

(- 2 p + 1)^{2} - 4 \cdot 1 \cdot (p^{2} - p - 20) = 9

x_{1, 2} = \frac{- ( - 2 p + 1 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 p + 1 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( - 2 p + 1 ) - 9}{2 \cdot 1}

x = \frac{- ( - 2 p + 1 ) + 9}{2 \cdot 1} : p + 4

x = \frac{- ( - 2 p + 1 ) - 9}{2 \cdot 1} : p - 5

The solutions to the quadratic equation are:

x = p + 4, x = p - 5

x^{2} + 5 x + 5 = 0

x, t^{2} = 240 x^{2}

12 x^{2} - 10 x + 1 = 7

900 x^{2} = 9

4 x^{2} = - 12 x

What are the solutions to the equation solvefor x,x^2-(2p-1)x+(p^2-p-20)=0 ?
The solutions to the equation solvefor x,x^2-(2p-1)x+(p^2-p-20)=0 are x=p+4,x=p-5
Find the zeros of solvefor x,x^2-(2p-1)x+(p^2-p-20)=0
The zeros of solvefor x,x^2-(2p-1)x+(p^2-p-20)=0 are x=p+4,x=p-5