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

The solutions to the equation solvefor p,2(p-q)^2-(p-q)-1=0 are p=q+1,p=(2q-1)/2

Find the zeros of solvefor p,2(p-q)^2-(p-q)-1=0

The zeros of solvefor p,2(p-q)^2-(p-q)-1=0 are p=q+1,p=(2q-1)/2

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solve for

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

p = q + 1, p = \frac{2 q - 1}{2}

Solution steps

2 (p - q)^{2} - (p - q) - 1 = 0

Expand

2 (p - q)^{2} - (p - q) - 1 : 2 p^{2} - 4 pq + 2 q^{2} - p + q - 1

2 p^{2} - 4 pq + 2 q^{2} - p + q - 1 = 0

Write in the standard form

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

2 p^{2} - (4 q + 1) p + 2 q^{2} + q - 1 = 0

Solve with the quadratic formula

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

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

p_{1, 2} = \frac{- ( - 4 q - 1 ) \pm 3}{2 \cdot 2}

Separate the solutions

p_{1} = \frac{- ( - 4 q - 1 ) + 3}{2 \cdot 2}, p_{2} = \frac{- ( - 4 q - 1 ) - 3}{2 \cdot 2}

p = \frac{- ( - 4 q - 1 ) + 3}{2 \cdot 2} : q + 1

p = \frac{- ( - 4 q - 1 ) - 3}{2 \cdot 2} : \frac{2 q - 1}{2}

The solutions to the quadratic equation are:

p = q + 1, p = \frac{2 q - 1}{2}

m^{2} + m - 56 = 0

m^{2} - 9 m + 20 = 0

x^{2} + 20 x + 85 = 2

z - z^{2} - \frac{1}{8} = 0

x, x^{2} - 36 = 0

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