What are the solutions to the equation x^2+68=-18x ?

The solutions to the equation x^2+68=-18x are x=-9+sqrt(13),x=-9-sqrt(13)

Find the zeros of x^2+68=-18x

The zeros of x^2+68=-18x are x=-9+sqrt(13),x=-9-sqrt(13)

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x^{2} + 68 = - 18 x

x = - 9 + 13, x = - 9 - 13

Decimal

x = - 5.39444 \dots, x = - 12.60555 \dots

Solution steps

x^{2} + 68 = - 18 x

Move

18 x

to the left side

x^{2} + 68 + 18 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + 18 x + 68 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 1 \cdot 68}{2 \cdot 1}

1 8^{2} - 4 \cdot 1 \cdot 68 = 213

x_{1, 2} = \frac{- 18 \pm 2 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 18 + 2 13}{2 \cdot 1}, x_{2} = \frac{- 18 - 2 13}{2 \cdot 1}

x = \frac{- 18 + 2 13}{2 \cdot 1} : - 9 + 13

x = \frac{- 18 - 2 13}{2 \cdot 1} : - 9 - 13

The solutions to the quadratic equation are:

x = - 9 + 13, x = - 9 - 13

2 x^{2} - 7 x + 5 = 10

5 r^{2} - 7 r + 4 = 6

x^{2} + 4 x + 29

5 (m + n - 1) x^{2} + 4 (m - n) x + \frac{mn}{12} = 0

x^{2} - 7 x - 144 = 0

What are the solutions to the equation x^2+68=-18x ?
The solutions to the equation x^2+68=-18x are x=-9+sqrt(13),x=-9-sqrt(13)
Find the zeros of x^2+68=-18x
The zeros of x^2+68=-18x are x=-9+sqrt(13),x=-9-sqrt(13)