What are the solutions to the equation p^2-3p=8 ?

The solutions to the equation p^2-3p=8 are p=(3+sqrt(41))/2 ,p=(3-sqrt(41))/2

Find the zeros of p^2-3p=8

The zeros of p^2-3p=8 are p=(3+sqrt(41))/2 ,p=(3-sqrt(41))/2

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p^{2} - 3 p = 8

p = \frac{3 + 41}{2}, p = \frac{3 - 41}{2}

Decimal Notation

p = 4.70156 \dots, p = - 1.70156 \dots

Solution steps

p^{2} - 3 p = 8

Move

8

to the left side

p^{2} - 3 p - 8 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 8) = 41

p_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 1}, p_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 1}

p = \frac{- ( - 3 ) + 41}{2 \cdot 1} : \frac{3 + 41}{2}

p = \frac{- ( - 3 ) - 41}{2 \cdot 1} : \frac{3 - 41}{2}

The solutions to the quadratic equation are:

p = \frac{3 + 41}{2}, p = \frac{3 - 41}{2}

(x - 2) (x - 6) = (x - 2) (x - 6)

4 (7 x - 2) + x = 4 x + 8

(x - y)^{2} - 16 y^{2}

\frac{1}{4} - \frac{1}{3}

2 ∣ x + 3 ∣ + 4 = 16

What are the solutions to the equation p^2-3p=8 ?
The solutions to the equation p^2-3p=8 are p=(3+sqrt(41))/2 ,p=(3-sqrt(41))/2
Find the zeros of p^2-3p=8
The zeros of p^2-3p=8 are p=(3+sqrt(41))/2 ,p=(3-sqrt(41))/2