What are the solutions to the equation -p^2=11p+18 ?

The solutions to the equation -p^2=11p+18 are p=-9,p=-2

Find the zeros of -p^2=11p+18

The zeros of -p^2=11p+18 are p=-9,p=-2

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- p^{2} = 11 p + 18

p = - 9, p = - 2

Solution steps

- p^{2} = 11 p + 18

Move

18

to the left side

- p^{2} - 18 = 11 p

Move

11 p

to the left side

- p^{2} - 18 - 11 p = 0

Write in the standard form

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

- p^{2} - 11 p - 18 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 ( - 1 ) ( - 18 )}{2 ( - 1 )}

(- 11)^{2} - 4 (- 1) (- 18) = 7

p_{1, 2} = \frac{- ( - 11 ) \pm 7}{2 ( - 1 )}

Separate the solutions

p_{1} = \frac{- ( - 11 ) + 7}{2 ( - 1 )}, p_{2} = \frac{- ( - 11 ) - 7}{2 ( - 1 )}

p = \frac{- ( - 11 ) + 7}{2 ( - 1 )} : - 9

p = \frac{- ( - 11 ) - 7}{2 ( - 1 )} : - 2

The solutions to the quadratic equation are:

p = - 9, p = - 2

(x + 13)^{2} - 441 = 0

x^{2} - 2 a x + 1 = 0

u^{2} + 9 = 0

x^{2} = 1 6^{2} - (83)^{2}

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

What are the solutions to the equation -p^2=11p+18 ?
The solutions to the equation -p^2=11p+18 are p=-9,p=-2
Find the zeros of -p^2=11p+18
The zeros of -p^2=11p+18 are p=-9,p=-2