What are the solutions to the equation 3p^2-13p+4=-4 ?

The solutions to the equation 3p^2-13p+4=-4 are p=(13+sqrt(73))/6 ,p=(13-sqrt(73))/6

Find the zeros of 3p^2-13p+4=-4

The zeros of 3p^2-13p+4=-4 are p=(13+sqrt(73))/6 ,p=(13-sqrt(73))/6

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3 p^{2} - 13 p + 4 = - 4

p = \frac{13 + 73}{6}, p = \frac{13 - 73}{6}

Decimal Notation

p = 3.59066 \dots, p = 0.74266 \dots

Solution steps

3 p^{2} - 13 p + 4 = - 4

Move

4

to the left side

3 p^{2} - 13 p + 8 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 3 \cdot 8}{2 \cdot 3}

(- 13)^{2} - 4 \cdot 3 \cdot 8 = 73

p_{1, 2} = \frac{- ( - 13 ) \pm 73}{2 \cdot 3}

Separate the solutions

p_{1} = \frac{- ( - 13 ) + 73}{2 \cdot 3}, p_{2} = \frac{- ( - 13 ) - 73}{2 \cdot 3}

p = \frac{- ( - 13 ) + 73}{2 \cdot 3} : \frac{13 + 73}{6}

p = \frac{- ( - 13 ) - 73}{2 \cdot 3} : \frac{13 - 73}{6}

The solutions to the quadratic equation are:

p = \frac{13 + 73}{6}, p = \frac{13 - 73}{6}

(x) (x + 1) = 240

2 x^{2} - x - 11 = - 3

5 x^{2} + 5 x - \frac{3}{4} = 0

3 x^{2} + 12 x = 7 x + 2

(x - 1)^{2}

What are the solutions to the equation 3p^2-13p+4=-4 ?
The solutions to the equation 3p^2-13p+4=-4 are p=(13+sqrt(73))/6 ,p=(13-sqrt(73))/6
Find the zeros of 3p^2-13p+4=-4
The zeros of 3p^2-13p+4=-4 are p=(13+sqrt(73))/6 ,p=(13-sqrt(73))/6