What are the solutions to the equation -5x^2-x=6 ?

The solutions to the equation -5x^2-x=6 are x=-1/10-i(sqrt(119))/(10),x=-1/10+i(sqrt(119))/(10)

Find the zeros of -5x^2-x=6

The zeros of -5x^2-x=6 are x=-1/10-i(sqrt(119))/(10),x=-1/10+i(sqrt(119))/(10)

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- 5 x^{2} - x = 6

x = - \frac{1}{10} - i \frac{119}{10}, x = - \frac{1}{10} + i \frac{119}{10}

Solution steps

- 5 x^{2} - x = 6

Move

6

to the left side

- 5 x^{2} - x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 5 ) ( - 6 )}{2 ( - 5 )}

Simplify

(- 1)^{2} - 4 (- 5) (- 6) : 119 i

x_{1, 2} = \frac{- ( - 1 ) \pm 119 i}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 119 i}{2 ( - 5 )}, x_{2} = \frac{- ( - 1 ) - 119 i}{2 ( - 5 )}

x = \frac{- ( - 1 ) + 119 i}{2 ( - 5 )} : - \frac{1}{10} - i \frac{119}{10}

x = \frac{- ( - 1 ) - 119 i}{2 ( - 5 )} : - \frac{1}{10} + i \frac{119}{10}

The solutions to the quadratic equation are:

x = - \frac{1}{10} - i \frac{119}{10}, x = - \frac{1}{10} + i \frac{119}{10}

6 n - (n + 1) < 3 n - 2 + 5 n

(\frac{6}{4})^{2}

\frac{1 2 ^{\frac{1}{4}}}{1 2 ^{\frac{3}{4}}}

\frac{x - 2}{5} - 1 = \frac{x - 9}{6}

2 r^{2} - 7 r + 6

What are the solutions to the equation -5x^2-x=6 ?
The solutions to the equation -5x^2-x=6 are x=-1/10-i(sqrt(119))/(10),x=-1/10+i(sqrt(119))/(10)
Find the zeros of -5x^2-x=6
The zeros of -5x^2-x=6 are x=-1/10-i(sqrt(119))/(10),x=-1/10+i(sqrt(119))/(10)