What are the solutions to the equation 3p^2-3p-6=-2 ?

The solutions to the equation 3p^2-3p-6=-2 are p=(3+sqrt(57))/6 ,p=(3-sqrt(57))/6

Find the zeros of 3p^2-3p-6=-2

The zeros of 3p^2-3p-6=-2 are p=(3+sqrt(57))/6 ,p=(3-sqrt(57))/6

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3 p^{2} - 3 p - 6 = - 2

p = \frac{3 + 57}{6}, p = \frac{3 - 57}{6}

Decimal Notation

p = 1.75830 \dots, p = - 0.75830 \dots

Solution steps

3 p^{2} - 3 p - 6 = - 2

Move

2

to the left side

3 p^{2} - 3 p - 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 3 (- 4) = 57

p_{1, 2} = \frac{- ( - 3 ) \pm 57}{2 \cdot 3}

Separate the solutions

p_{1} = \frac{- ( - 3 ) + 57}{2 \cdot 3}, p_{2} = \frac{- ( - 3 ) - 57}{2 \cdot 3}

p = \frac{- ( - 3 ) + 57}{2 \cdot 3} : \frac{3 + 57}{6}

p = \frac{- ( - 3 ) - 57}{2 \cdot 3} : \frac{3 - 57}{6}

The solutions to the quadratic equation are:

p = \frac{3 + 57}{6}, p = \frac{3 - 57}{6}

64 x^{2} - 212 x + 49 = 0

x^{2} = 4 x - 29

b^{2} + b - 1 = 0

0 = 12 x^{2} - 12 x

2 = (\frac{1}{2}) (x - 4)^{2}

What are the solutions to the equation 3p^2-3p-6=-2 ?
The solutions to the equation 3p^2-3p-6=-2 are p=(3+sqrt(57))/6 ,p=(3-sqrt(57))/6
Find the zeros of 3p^2-3p-6=-2
The zeros of 3p^2-3p-6=-2 are p=(3+sqrt(57))/6 ,p=(3-sqrt(57))/6