What are the solutions to the equation 6c^2+9=5c ?

The solutions to the equation 6c^2+9=5c are c= 5/12+i(sqrt(191))/(12),c= 5/12-i(sqrt(191))/(12)

Find the zeros of 6c^2+9=5c

The zeros of 6c^2+9=5c are c= 5/12+i(sqrt(191))/(12),c= 5/12-i(sqrt(191))/(12)

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6 c^{2} + 9 = 5 c

c = \frac{5}{12} + i \frac{191}{12}, c = \frac{5}{12} - i \frac{191}{12}

Solution steps

6 c^{2} + 9 = 5 c

Move

5 c

to the left side

6 c^{2} + 9 - 5 c = 0

Write in the standard form

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

6 c^{2} - 5 c + 9 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 9}{2 \cdot 6}

Simplify

(- 5)^{2} - 4 \cdot 6 \cdot 9 : 191 i

c_{1, 2} = \frac{- ( - 5 ) \pm 191 i}{2 \cdot 6}

Separate the solutions

c_{1} = \frac{- ( - 5 ) + 191 i}{2 \cdot 6}, c_{2} = \frac{- ( - 5 ) - 191 i}{2 \cdot 6}

c = \frac{- ( - 5 ) + 191 i}{2 \cdot 6} : \frac{5}{12} + i \frac{191}{12}

c = \frac{- ( - 5 ) - 191 i}{2 \cdot 6} : \frac{5}{12} - i \frac{191}{12}

The solutions to the quadratic equation are:

c = \frac{5}{12} + i \frac{191}{12}, c = \frac{5}{12} - i \frac{191}{12}

8 x^{2} + 6 x = 8

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

- x^{2} + 6 x + 12 = 0

0.000015 (x + 100)^{2} = 0.2254

v^{2} - 3 v - 10 = 0

What are the solutions to the equation 6c^2+9=5c ?
The solutions to the equation 6c^2+9=5c are c= 5/12+i(sqrt(191))/(12),c= 5/12-i(sqrt(191))/(12)
Find the zeros of 6c^2+9=5c
The zeros of 6c^2+9=5c are c= 5/12+i(sqrt(191))/(12),c= 5/12-i(sqrt(191))/(12)