What are the solutions to the equation p^2+7p-8=0 ?

The solutions to the equation p^2+7p-8=0 are p=1,p=-8

Find the zeros of p^2+7p-8=0

The zeros of p^2+7p-8=0 are p=1,p=-8

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p^{2} + 7 p - 8 = 0

p = 1, p = - 8

Solution steps

p^{2} + 7 p - 8 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 8) = 9

p_{1, 2} = \frac{- 7 \pm 9}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 7 + 9}{2 \cdot 1}, p_{2} = \frac{- 7 - 9}{2 \cdot 1}

p = \frac{- 7 + 9}{2 \cdot 1} : 1

p = \frac{- 7 - 9}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

p = 1, p = - 8

y, x^{2} + 2 x y - 4 + y^{2} = 0

10 x^{2} + 15 x = 0

26 x^{2} + 25 x - 975 = 0

r^{2} + 15 r + 50 = 6 - r

(p + 1)^{2} = \frac{4}{9}

What are the solutions to the equation p^2+7p-8=0 ?
The solutions to the equation p^2+7p-8=0 are p=1,p=-8
Find the zeros of p^2+7p-8=0
The zeros of p^2+7p-8=0 are p=1,p=-8