What are the solutions to the equation 2p^2-3p-7=0 ?

The solutions to the equation 2p^2-3p-7=0 are p=(3+sqrt(65))/4 ,p=(3-sqrt(65))/4

Find the zeros of 2p^2-3p-7=0

The zeros of 2p^2-3p-7=0 are p=(3+sqrt(65))/4 ,p=(3-sqrt(65))/4

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2 p^{2} - 3 p - 7 = 0

p = \frac{3 + 65}{4}, p = \frac{3 - 65}{4}

Decimal Notation

p = 2.76556 \dots, p = - 1.26556 \dots

Solution steps

2 p^{2} - 3 p - 7 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 7) = 65

p_{1, 2} = \frac{- ( - 3 ) \pm 65}{2 \cdot 2}

Separate the solutions

p_{1} = \frac{- ( - 3 ) + 65}{2 \cdot 2}, p_{2} = \frac{- ( - 3 ) - 65}{2 \cdot 2}

p = \frac{- ( - 3 ) + 65}{2 \cdot 2} : \frac{3 + 65}{4}

p = \frac{- ( - 3 ) - 65}{2 \cdot 2} : \frac{3 - 65}{4}

The solutions to the quadratic equation are:

p = \frac{3 + 65}{4}, p = \frac{3 - 65}{4}

(\frac{1}{9})^{3 t + 5} \geq (\frac{1}{243})^{t - 6}

x^{3} + 7 x^{2} + 14 x + 8

6.5 (x - 4) = 19.5

2 n^{2} + 5 n - 1 < 0

(x + 4)^{2} = x^{2} + 8

What are the solutions to the equation 2p^2-3p-7=0 ?
The solutions to the equation 2p^2-3p-7=0 are p=(3+sqrt(65))/4 ,p=(3-sqrt(65))/4
Find the zeros of 2p^2-3p-7=0
The zeros of 2p^2-3p-7=0 are p=(3+sqrt(65))/4 ,p=(3-sqrt(65))/4