What are the solutions to the equation x^2+2x-80=0 ?

The solutions to the equation x^2+2x-80=0 are x=8,x=-10

Find the zeros of x^2+2x-80=0

The zeros of x^2+2x-80=0 are x=8,x=-10

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x^{2} + 2 x - 80 = 0

x = 8, x = - 10

Solution steps

x^{2} + 2 x - 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 80 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 80) = 18

x_{1, 2} = \frac{- 2 \pm 18}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 18}{2 \cdot 1}, x_{2} = \frac{- 2 - 18}{2 \cdot 1}

x = \frac{- 2 + 18}{2 \cdot 1} : 8

x = \frac{- 2 - 18}{2 \cdot 1} : - 10

The solutions to the quadratic equation are:

x = 8, x = - 10

\frac{- 4 p q ^{5} r ^{3}}{8 p ^{2} q ^{2} r ^{10}}

\frac{5}{8} - \frac{6}{8}

s^{2} + 6 s + 25 = 0

1 < 3

- \frac{1}{4} - 4

What are the solutions to the equation x^2+2x-80=0 ?
The solutions to the equation x^2+2x-80=0 are x=8,x=-10
Find the zeros of x^2+2x-80=0
The zeros of x^2+2x-80=0 are x=8,x=-10