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

The solutions to the equation 6y^2-5y+2=0 are y= 5/12+i(sqrt(23))/(12),y= 5/12-i(sqrt(23))/(12)

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

The zeros of 6y^2-5y+2=0 are y= 5/12+i(sqrt(23))/(12),y= 5/12-i(sqrt(23))/(12)

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

y = \frac{5}{12} + i \frac{23}{12}, y = \frac{5}{12} - i \frac{23}{12}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 6 \cdot 2 : 23 i

y_{1, 2} = \frac{- ( - 5 ) \pm 23 i}{2 \cdot 6}

Separate the solutions

y_{1} = \frac{- ( - 5 ) + 23 i}{2 \cdot 6}, y_{2} = \frac{- ( - 5 ) - 23 i}{2 \cdot 6}

y = \frac{- ( - 5 ) + 23 i}{2 \cdot 6} : \frac{5}{12} + i \frac{23}{12}

y = \frac{- ( - 5 ) - 23 i}{2 \cdot 6} : \frac{5}{12} - i \frac{23}{12}

The solutions to the quadratic equation are:

y = \frac{5}{12} + i \frac{23}{12}, y = \frac{5}{12} - i \frac{23}{12}

- 3 x^{2} = 3

x^{2} - 12 x + 7 = 0

4 r^{3} s + 864 s^{4} = 4 s (r + 6 s)

(x - 14)^{2} + 13 = 12

ab x^{2} = (a - b)^{2} (x + 1)

What are the solutions to the equation 6y^2-5y+2=0 ?
The solutions to the equation 6y^2-5y+2=0 are y= 5/12+i(sqrt(23))/(12),y= 5/12-i(sqrt(23))/(12)
Find the zeros of 6y^2-5y+2=0
The zeros of 6y^2-5y+2=0 are y= 5/12+i(sqrt(23))/(12),y= 5/12-i(sqrt(23))/(12)