What are the solutions to the equation 6x^2-5x+6=0 ?

The solutions to the equation 6x^2-5x+6=0 are x= 5/12+i(sqrt(119))/(12),x= 5/12-i(sqrt(119))/(12)

Find the zeros of 6x^2-5x+6=0

The zeros of 6x^2-5x+6=0 are x= 5/12+i(sqrt(119))/(12),x= 5/12-i(sqrt(119))/(12)

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

x = \frac{5}{12} + i \frac{119}{12}, x = \frac{5}{12} - i \frac{119}{12}

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

x_{1, 2} = \frac{- ( - 5 ) \pm 119 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 119 i}{2 \cdot 6}, x_{2} = \frac{- ( - 5 ) - 119 i}{2 \cdot 6}

x = \frac{- ( - 5 ) + 119 i}{2 \cdot 6} : \frac{5}{12} + i \frac{119}{12}

x = \frac{- ( - 5 ) - 119 i}{2 \cdot 6} : \frac{5}{12} - i \frac{119}{12}

The solutions to the quadratic equation are:

x = \frac{5}{12} + i \frac{119}{12}, x = \frac{5}{12} - i \frac{119}{12}

∣ x + 2 ∣ \leq 4

\frac{3}{4} q - \frac{2}{3} q < \frac{5}{12} q - \frac{1}{2}

5200

2 x + 2 = 10

9 x^{2} + 8 x - 1 = 0

What are the solutions to the equation 6x^2-5x+6=0 ?
The solutions to the equation 6x^2-5x+6=0 are x= 5/12+i(sqrt(119))/(12),x= 5/12-i(sqrt(119))/(12)
Find the zeros of 6x^2-5x+6=0
The zeros of 6x^2-5x+6=0 are x= 5/12+i(sqrt(119))/(12),x= 5/12-i(sqrt(119))/(12)