What are the solutions to the equation solvefor p,I=40p-p^2 ?

The solutions to the equation solvefor p,I=40p-p^2 are p=20-sqrt(-I+400),p=sqrt(-I+400)+20

Find the zeros of solvefor p,I=40p-p^2

The zeros of solvefor p,I=40p-p^2 are p=20-sqrt(-I+400),p=sqrt(-I+400)+20

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solve for

p, I = 40 p - p^{2}

p = 20 - - I + 400, p = - I + 400 + 20

Solution steps

I = 40 p - p^{2}

Switch sides

40 p - p^{2} = I

Move

I

to the left side

40 p - p^{2} - I = 0

Write in the standard form

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

- p^{2} + 40 p - I = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 1 ) ( - I )}{2 ( - 1 )}

Simplify

4 0^{2} - 4 (- 1) (- I) : 2 400 - I

p_{1, 2} = \frac{- 40 \pm 2 400 - I}{2 ( - 1 )}

Separate the solutions

p_{1} = \frac{- 40 + 2 400 - I}{2 ( - 1 )}, p_{2} = \frac{- 40 - 2 400 - I}{2 ( - 1 )}

p = \frac{- 40 + 2 400 - I}{2 ( - 1 )} : 20 - - I + 400

p = \frac{- 40 - 2 400 - I}{2 ( - 1 )} : - I + 400 + 20

The solutions to the quadratic equation are:

p = 20 - - I + 400, p = - I + 400 + 20

y, z = 9 - y^{2}

3 x (x - 8) + 2 (x - 7) = 2

- 8 x^{2} + 2 x = 5

y, z = x^{3} + 4 x y - 3 y^{2}

x, x^{2} - 3 x y + 5 y^{2} - 4 = 0

What are the solutions to the equation solvefor p,I=40p-p^2 ?
The solutions to the equation solvefor p,I=40p-p^2 are p=20-sqrt(-I+400),p=sqrt(-I+400)+20
Find the zeros of solvefor p,I=40p-p^2
The zeros of solvefor p,I=40p-p^2 are p=20-sqrt(-I+400),p=sqrt(-I+400)+20