What are the solutions to the equation 2x^2+39=-18x ?

The solutions to the equation 2x^2+39=-18x are x=(-9+sqrt(3))/2 ,x=-(9+sqrt(3))/2

Find the zeros of 2x^2+39=-18x

The zeros of 2x^2+39=-18x are x=(-9+sqrt(3))/2 ,x=-(9+sqrt(3))/2

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2 x^{2} + 39 = - 18 x

x = \frac{- 9 + 3}{2}, x = - \frac{9 + 3}{2}

Decimal Notation

x = - 3.63397 \dots, x = - 5.36602 \dots

Solution steps

2 x^{2} + 39 = - 18 x

Move

18 x

to the left side

2 x^{2} + 39 + 18 x = 0

Write in the standard form

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

2 x^{2} + 18 x + 39 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 2 \cdot 39}{2 \cdot 2}

1 8^{2} - 4 \cdot 2 \cdot 39 = 23

x_{1, 2} = \frac{- 18 \pm 2 3}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 18 + 2 3}{2 \cdot 2}, x_{2} = \frac{- 18 - 2 3}{2 \cdot 2}

x = \frac{- 18 + 2 3}{2 \cdot 2} : \frac{- 9 + 3}{2}

x = \frac{- 18 - 2 3}{2 \cdot 2} : - \frac{9 + 3}{2}

The solutions to the quadratic equation are:

x = \frac{- 9 + 3}{2}, x = - \frac{9 + 3}{2}

\frac{10 i}{3 + i}

- x^{2} + 8 x - 15 = 0

x > - 2 and x < 11

e^{x} - 2 < 0

2 x = x + 5

What are the solutions to the equation 2x^2+39=-18x ?
The solutions to the equation 2x^2+39=-18x are x=(-9+sqrt(3))/2 ,x=-(9+sqrt(3))/2
Find the zeros of 2x^2+39=-18x
The zeros of 2x^2+39=-18x are x=(-9+sqrt(3))/2 ,x=-(9+sqrt(3))/2