What are the solutions to the equation 2p^2-3=11p ?

The solutions to the equation 2p^2-3=11p are p=(11+sqrt(145))/4 ,p=(11-sqrt(145))/4

Find the zeros of 2p^2-3=11p

The zeros of 2p^2-3=11p are p=(11+sqrt(145))/4 ,p=(11-sqrt(145))/4

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2 p^{2} - 3 = 11 p

p = \frac{11 + 145}{4}, p = \frac{11 - 145}{4}

Decimal Notation

p = 5.76039 \dots, p = - 0.26039 \dots

Solution steps

2 p^{2} - 3 = 11 p

Move

11 p

to the left side

2 p^{2} - 3 - 11 p = 0

Write in the standard form

a x^{2} + b x + c = 0

2 p^{2} - 11 p - 3 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

(- 11)^{2} - 4 \cdot 2 (- 3) = 145

p_{1, 2} = \frac{- ( - 11 ) \pm 145}{2 \cdot 2}

Separate the solutions

p_{1} = \frac{- ( - 11 ) + 145}{2 \cdot 2}, p_{2} = \frac{- ( - 11 ) - 145}{2 \cdot 2}

p = \frac{- ( - 11 ) + 145}{2 \cdot 2} : \frac{11 + 145}{4}

p = \frac{- ( - 11 ) - 145}{2 \cdot 2} : \frac{11 - 145}{4}

The solutions to the quadratic equation are:

p = \frac{11 + 145}{4}, p = \frac{11 - 145}{4}

3 x^{2} - 6 x - 72 = 0

\frac{1}{2} x + \frac{1}{3} x - 7 = \frac{5}{6} x

\frac{3 π}{2} (\frac{180}{π})

x^{3} - 7 x^{2} + 17 x - 20

- (- 10) = a - [5 + 3 a - {5 a - (a + 6)}] + 13

What are the solutions to the equation 2p^2-3=11p ?
The solutions to the equation 2p^2-3=11p are p=(11+sqrt(145))/4 ,p=(11-sqrt(145))/4
Find the zeros of 2p^2-3=11p
The zeros of 2p^2-3=11p are p=(11+sqrt(145))/4 ,p=(11-sqrt(145))/4