What are the solutions to the equation p+45=p(p-3) ?

The solutions to the equation p+45=p(p-3) are p=9,p=-5

Find the zeros of p+45=p(p-3)

The zeros of p+45=p(p-3) are p=9,p=-5

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p + 45 = p (p - 3)

p = 9, p = - 5

Solution steps

p + 45 = p (p - 3)

Expand

p (p - 3) : p^{2} - 3 p

p + 45 = p^{2} - 3 p

Switch sides

p^{2} - 3 p = p + 45

Move

45

to the left side

p^{2} - 3 p - 45 = p

Move

p

to the left side

p^{2} - 4 p - 45 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 45 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 45) = 14

p_{1, 2} = \frac{- ( - 4 ) \pm 14}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- ( - 4 ) + 14}{2 \cdot 1}, p_{2} = \frac{- ( - 4 ) - 14}{2 \cdot 1}

p = \frac{- ( - 4 ) + 14}{2 \cdot 1} : 9

p = \frac{- ( - 4 ) - 14}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

p = 9, p = - 5

x^{3} + 4 x^{2} - 3 x - 14

\frac{( \frac{1}{4} + \frac{1}{2} )}{2}

i^{95}

4 c = 3

\frac{8 x ^{2} y}{8 x + y} \cdot \frac{x + y}{11 x}

What are the solutions to the equation p+45=p(p-3) ?
The solutions to the equation p+45=p(p-3) are p=9,p=-5
Find the zeros of p+45=p(p-3)
The zeros of p+45=p(p-3) are p=9,p=-5