What are the solutions to the equation -2p^2+7p-4=-3p^2 ?

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

Find the zeros of -2p^2+7p-4=-3p^2

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

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- 2 p^{2} + 7 p - 4 = - 3 p^{2}

p = \frac{- 7 + 65}{2}, p = \frac{- 7 - 65}{2}

Decimal Notation

p = 0.53112 \dots, p = - 7.53112 \dots

Solution steps

- 2 p^{2} + 7 p - 4 = - 3 p^{2}

Move

3 p^{2}

to the left side

p^{2} + 7 p - 4 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 1 \cdot (- 4) = 65

p_{1, 2} = \frac{- 7 \pm 65}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 7 + 65}{2 \cdot 1}, p_{2} = \frac{- 7 - 65}{2 \cdot 1}

p = \frac{- 7 + 65}{2 \cdot 1} : \frac{- 7 + 65}{2}

p = \frac{- 7 - 65}{2 \cdot 1} : \frac{- 7 - 65}{2}

The solutions to the quadratic equation are:

p = \frac{- 7 + 65}{2}, p = \frac{- 7 - 65}{2}

(x - 8)^{2} - 2 x = 7

3 x^{2} + 5 x = 12

25 b^{2} - 16 = 0

3.5 x^{2} - 5 x - 2.5 = 0

- 8 x = x^{2} - 12

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