What are the solutions to the equation x^2+22x+120.5=0 ?

The solutions to the equation x^2+22x+120.5=0 are x=(-22+sqrt(2))/2 ,x=-(22+sqrt(2))/2

Find the zeros of x^2+22x+120.5=0

The zeros of x^2+22x+120.5=0 are x=(-22+sqrt(2))/2 ,x=-(22+sqrt(2))/2

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x^{2} + 22 x + 120.5 = 0

x = \frac{- 22 + 2}{2}, x = - \frac{22 + 2}{2}

Decimal Notation

x = - 10.29289 \dots, x = - 11.70710 \dots

Solution steps

x^{2} + 22 x + 120.5 = 0

Multiply both sides by

10

10 x^{2} + 220 x + 1205 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 220 \pm 22 0 ^{2} - 4 \cdot 10 \cdot 1205}{2 \cdot 10}

22 0^{2} - 4 \cdot 10 \cdot 1205 = 102

x_{1, 2} = \frac{- 220 \pm 10 2}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- 220 + 10 2}{2 \cdot 10}, x_{2} = \frac{- 220 - 10 2}{2 \cdot 10}

x = \frac{- 220 + 10 2}{2 \cdot 10} : \frac{- 22 + 2}{2}

x = \frac{- 220 - 10 2}{2 \cdot 10} : - \frac{22 + 2}{2}

The solutions to the quadratic equation are:

x = \frac{- 22 + 2}{2}, x = - \frac{22 + 2}{2}

x^{2} - 9 x + c = 0

x^{2} = 0.09

x^{2} - 4 x - 21 = 0

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

t, h = \frac{1}{2} g t^{2}

What are the solutions to the equation x^2+22x+120.5=0 ?
The solutions to the equation x^2+22x+120.5=0 are x=(-22+sqrt(2))/2 ,x=-(22+sqrt(2))/2
Find the zeros of x^2+22x+120.5=0
The zeros of x^2+22x+120.5=0 are x=(-22+sqrt(2))/2 ,x=-(22+sqrt(2))/2