What are the solutions to the equation p^2-25p+39=-5p ?

The solutions to the equation p^2-25p+39=-5p are p=10+sqrt(61),p=10-sqrt(61)

Find the zeros of p^2-25p+39=-5p

The zeros of p^2-25p+39=-5p are p=10+sqrt(61),p=10-sqrt(61)

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p^{2} - 25 p + 39 = - 5 p

p = 10 + 61, p = 10 - 61

Decimal Notation

p = 17.81024 \dots, p = 2.18975 \dots

Solution steps

p^{2} - 25 p + 39 = - 5 p

Move

5 p

to the left side

p^{2} - 20 p + 39 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 1 \cdot 39}{2 \cdot 1}

(- 20)^{2} - 4 \cdot 1 \cdot 39 = 261

p_{1, 2} = \frac{- ( - 20 ) \pm 2 61}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- ( - 20 ) + 2 61}{2 \cdot 1}, p_{2} = \frac{- ( - 20 ) - 2 61}{2 \cdot 1}

p = \frac{- ( - 20 ) + 2 61}{2 \cdot 1} : 10 + 61

p = \frac{- ( - 20 ) - 2 61}{2 \cdot 1} : 10 - 61

The solutions to the quadratic equation are:

p = 10 + 61, p = 10 - 61

5 x^{2} - 10 x - 5 = 0

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

(x + 1) 3 - (x - 1) 3 = 6 x (x - 3)

5 x^{2} + 14 x + 2 = 0

x^{2} - 7 x + 31 = 0

What are the solutions to the equation p^2-25p+39=-5p ?
The solutions to the equation p^2-25p+39=-5p are p=10+sqrt(61),p=10-sqrt(61)
Find the zeros of p^2-25p+39=-5p
The zeros of p^2-25p+39=-5p are p=10+sqrt(61),p=10-sqrt(61)