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 \cdot 5 \cdot 6}{2 \cdot 5}

Simplify

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

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

Separate the solutions

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

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

x = \frac{- ( - 1 ) - 119 i}{2 \cdot 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}

t^{2} + 10 t + 9 = 0

g^{2} - 10 g + 16 = 0

25 = 5 + 37 t - 16 t^{2}

y (y + 2) = 48

3 y^{2} - 2 y - 1 = 0

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)