What are the solutions to the equation x^2+20x=-68 ?

The solutions to the equation x^2+20x=-68 are x=2(2sqrt(2)-5),x=-2(5+2sqrt(2))

Find the zeros of x^2+20x=-68

The zeros of x^2+20x=-68 are x=2(2sqrt(2)-5),x=-2(5+2sqrt(2))

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x^{2} + 20 x = - 68

x = 2 (22 - 5), x = - 2 (5 + 22)

Decimal Notation

x = - 4.34314 \dots, x = - 15.65685 \dots

Solution steps

x^{2} + 20 x = - 68

Move

68

to the left side

x^{2} + 20 x + 68 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 68}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot 68 = 82

x_{1, 2} = \frac{- 20 \pm 8 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 8 2}{2 \cdot 1}, x_{2} = \frac{- 20 - 8 2}{2 \cdot 1}

x = \frac{- 20 + 8 2}{2 \cdot 1} : 2 (22 - 5)

x = \frac{- 20 - 8 2}{2 \cdot 1} : - 2 (5 + 22)

The solutions to the quadratic equation are:

x = 2 (22 - 5), x = - 2 (5 + 22)

\frac{x}{2} - 2 = \frac{4 x}{3} + 8

q - 12 = 1

x^{2} - 2 x + 11 = 0

2 x + 3 (0) = 5 x - 0

- 1 < \frac{3 x + 1}{2} \leq 5

What are the solutions to the equation x^2+20x=-68 ?
The solutions to the equation x^2+20x=-68 are x=2(2sqrt(2)-5),x=-2(5+2sqrt(2))
Find the zeros of x^2+20x=-68
The zeros of x^2+20x=-68 are x=2(2sqrt(2)-5),x=-2(5+2sqrt(2))