What are the solutions to the equation 6x^2-5x+3=0 ?

The solutions to the equation 6x^2-5x+3=0 are x= 5/12+i(sqrt(47))/(12),x= 5/12-i(sqrt(47))/(12)

Find the zeros of 6x^2-5x+3=0

The zeros of 6x^2-5x+3=0 are x= 5/12+i(sqrt(47))/(12),x= 5/12-i(sqrt(47))/(12)

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6 x^{2} - 5 x + 3 = 0

x = \frac{5}{12} + i \frac{47}{12}, x = \frac{5}{12} - i \frac{47}{12}

Solution steps

6 x^{2} - 5 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

Simplify

(- 5)^{2} - 4 \cdot 6 \cdot 3 : 47 i

x_{1, 2} = \frac{- ( - 5 ) \pm 47 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 47 i}{2 \cdot 6}, x_{2} = \frac{- ( - 5 ) - 47 i}{2 \cdot 6}

x = \frac{- ( - 5 ) + 47 i}{2 \cdot 6} : \frac{5}{12} + i \frac{47}{12}

x = \frac{- ( - 5 ) - 47 i}{2 \cdot 6} : \frac{5}{12} - i \frac{47}{12}

The solutions to the quadratic equation are:

x = \frac{5}{12} + i \frac{47}{12}, x = \frac{5}{12} - i \frac{47}{12}

\frac{( 5 x ^{4} y ^{3} )}{8 x ^{5}} \cdot \frac{( 3 x ^{3} y ^{5} )}{6 y ^{4}}

3 (x - 2) > 15

\frac{3 x - 5}{2} = 6

(24 c)^{\frac{2}{3}}

5 = \frac{3 x + 19}{2}

What are the solutions to the equation 6x^2-5x+3=0 ?
The solutions to the equation 6x^2-5x+3=0 are x= 5/12+i(sqrt(47))/(12),x= 5/12-i(sqrt(47))/(12)
Find the zeros of 6x^2-5x+3=0
The zeros of 6x^2-5x+3=0 are x= 5/12+i(sqrt(47))/(12),x= 5/12-i(sqrt(47))/(12)