What are the solutions to the equation 2p^2+3p-4=0 ?

The solutions to the equation 2p^2+3p-4=0 are p=(-3+sqrt(41))/4 ,p=(-3-sqrt(41))/4

Find the zeros of 2p^2+3p-4=0

The zeros of 2p^2+3p-4=0 are p=(-3+sqrt(41))/4 ,p=(-3-sqrt(41))/4

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

p = \frac{- 3 + 41}{4}, p = \frac{- 3 - 41}{4}

Decimal Notation

p = 0.85078 \dots, p = - 2.35078 \dots

Solution steps

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 2 (- 4) = 41

p_{1, 2} = \frac{- 3 \pm 41}{2 \cdot 2}

Separate the solutions

p_{1} = \frac{- 3 + 41}{2 \cdot 2}, p_{2} = \frac{- 3 - 41}{2 \cdot 2}

p = \frac{- 3 + 41}{2 \cdot 2} : \frac{- 3 + 41}{4}

p = \frac{- 3 - 41}{2 \cdot 2} : \frac{- 3 - 41}{4}

The solutions to the quadratic equation are:

p = \frac{- 3 + 41}{4}, p = \frac{- 3 - 41}{4}

x^{2} - (k + 3) x + 3 k = 0

x^{2} - x = - 4

x^{2} - 16 x + 9 = 0

24 x^{2} + 6 = 24 x

x, f (x + 1) = 3 x^{2} - 4 x + 1

What are the solutions to the equation 2p^2+3p-4=0 ?
The solutions to the equation 2p^2+3p-4=0 are p=(-3+sqrt(41))/4 ,p=(-3-sqrt(41))/4
Find the zeros of 2p^2+3p-4=0
The zeros of 2p^2+3p-4=0 are p=(-3+sqrt(41))/4 ,p=(-3-sqrt(41))/4