What are the solutions to the equation 6x^2=5x-4 ?

The solutions to the equation 6x^2=5x-4 are x= 5/12+i(sqrt(71))/(12),x= 5/12-i(sqrt(71))/(12)

Find the zeros of 6x^2=5x-4

The zeros of 6x^2=5x-4 are x= 5/12+i(sqrt(71))/(12),x= 5/12-i(sqrt(71))/(12)

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6 x^{2} = 5 x - 4

x = \frac{5}{12} + i \frac{71}{12}, x = \frac{5}{12} - i \frac{71}{12}

Solution steps

6 x^{2} = 5 x - 4

Move

4

to the left side

6 x^{2} + 4 = 5 x

Move

5 x

to the left side

6 x^{2} + 4 - 5 x = 0

Write in the standard form

a x^{2} + b x + c = 0

6 x^{2} - 5 x + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 6 \cdot 4 : 71 i

x_{1, 2} = \frac{- ( - 5 ) \pm 71 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 71 i}{2 \cdot 6}, x_{2} = \frac{- ( - 5 ) - 71 i}{2 \cdot 6}

x = \frac{- ( - 5 ) + 71 i}{2 \cdot 6} : \frac{5}{12} + i \frac{71}{12}

x = \frac{- ( - 5 ) - 71 i}{2 \cdot 6} : \frac{5}{12} - i \frac{71}{12}

The solutions to the quadratic equation are:

x = \frac{5}{12} + i \frac{71}{12}, x = \frac{5}{12} - i \frac{71}{12}

\frac{3}{4} x^{2} - \frac{1}{3} = 5

x^{2} - 11 x + 24 = 0

6 (y - 8)^{2} = 18

x^{2} + 6 x + 13 = 5

x^{2} + 18 x + 42

What are the solutions to the equation 6x^2=5x-4 ?
The solutions to the equation 6x^2=5x-4 are x= 5/12+i(sqrt(71))/(12),x= 5/12-i(sqrt(71))/(12)
Find the zeros of 6x^2=5x-4
The zeros of 6x^2=5x-4 are x= 5/12+i(sqrt(71))/(12),x= 5/12-i(sqrt(71))/(12)