What are the solutions to the equation x^2+22x+112=0 ?

The solutions to the equation x^2+22x+112=0 are x=-8,x=-14

Find the zeros of x^2+22x+112=0

The zeros of x^2+22x+112=0 are x=-8,x=-14

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x^{2} + 22 x + 112 = 0

x = - 8, x = - 14

Solution steps

x^{2} + 22 x + 112 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 \cdot 1 \cdot 112}{2 \cdot 1}

2 2^{2} - 4 \cdot 1 \cdot 112 = 6

x_{1, 2} = \frac{- 22 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 22 + 6}{2 \cdot 1}, x_{2} = \frac{- 22 - 6}{2 \cdot 1}

x = \frac{- 22 + 6}{2 \cdot 1} : - 8

x = \frac{- 22 - 6}{2 \cdot 1} : - 14

The solutions to the quadratic equation are:

x = - 8, x = - 14

8 1^{- \frac{1}{4}}

\frac{π}{8} + \frac{π}{8}

m^{2} - mn

x^{3} - 13 x + 12

\frac{1}{2} x - 5 = 17

What are the solutions to the equation x^2+22x+112=0 ?
The solutions to the equation x^2+22x+112=0 are x=-8,x=-14
Find the zeros of x^2+22x+112=0
The zeros of x^2+22x+112=0 are x=-8,x=-14