What are the solutions to the equation p^2+4p-12=0 ?

The solutions to the equation p^2+4p-12=0 are p=2,p=-6

Find the zeros of p^2+4p-12=0

The zeros of p^2+4p-12=0 are p=2,p=-6

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p^{2} + 4 p - 12 = 0

p = 2, p = - 6

Solution steps

p^{2} + 4 p - 12 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 12) = 8

p_{1, 2} = \frac{- 4 \pm 8}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 4 + 8}{2 \cdot 1}, p_{2} = \frac{- 4 - 8}{2 \cdot 1}

p = \frac{- 4 + 8}{2 \cdot 1} : 2

p = \frac{- 4 - 8}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

p = 2, p = - 6

y (2 + y) - y - 2

(\frac{0.03}{1})^{\frac{1}{2}} \cdot 0.5 \cdot 109

\frac{x + 5}{x - 7} \leq 1

(\frac{3 x}{y})^{2} (\frac{2 y}{3 x})^{3}

3 x^{2} - 8 x = 1

What are the solutions to the equation p^2+4p-12=0 ?
The solutions to the equation p^2+4p-12=0 are p=2,p=-6
Find the zeros of p^2+4p-12=0
The zeros of p^2+4p-12=0 are p=2,p=-6