What are the solutions to the equation x^2+2x-120=0 ?

The solutions to the equation x^2+2x-120=0 are x=10,x=-12

Find the zeros of x^2+2x-120=0

The zeros of x^2+2x-120=0 are x=10,x=-12

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x^{2} + 2 x - 120 = 0

x = 10, x = - 12

Solution steps

x^{2} + 2 x - 120 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 120 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 120) = 22

x_{1, 2} = \frac{- 2 \pm 22}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 22}{2 \cdot 1}, x_{2} = \frac{- 2 - 22}{2 \cdot 1}

x = \frac{- 2 + 22}{2 \cdot 1} : 10

x = \frac{- 2 - 22}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

x = 10, x = - 12

w^{3} + 27

- x^{2} - 2 x + 3 \leq 0

x - 8 = - 10

lo g_{10} (16)

\frac{6}{54}

What are the solutions to the equation x^2+2x-120=0 ?
The solutions to the equation x^2+2x-120=0 are x=10,x=-12
Find the zeros of x^2+2x-120=0
The zeros of x^2+2x-120=0 are x=10,x=-12