What are the solutions to the equation x^2+22x+120=0 ?

The solutions to the equation x^2+22x+120=0 are x=-10,x=-12

Find the zeros of x^2+22x+120=0

The zeros of x^2+22x+120=0 are x=-10,x=-12

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x^{2} + 22 x + 120 = 0

x = - 10, x = - 12

Solution steps

x^{2} + 22 x + 120 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 \cdot 1 \cdot 120}{2 \cdot 1}

2 2^{2} - 4 \cdot 1 \cdot 120 = 2

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 10, x = - 12

x^{2} + 2 x - 5

x^{2} + 2 x - 8

(x - 1)^{2} = - 8

x^{2} + 2 x = 0

(x - 1)^{2} = 81

What are the solutions to the equation x^2+22x+120=0 ?
The solutions to the equation x^2+22x+120=0 are x=-10,x=-12
Find the zeros of x^2+22x+120=0
The zeros of x^2+22x+120=0 are x=-10,x=-12