What are the solutions to the equation 12x-5=9x^2 ?

The solutions to the equation 12x-5=9x^2 are x= 2/3+i 1/3 ,x= 2/3-i 1/3

Find the zeros of 12x-5=9x^2

The zeros of 12x-5=9x^2 are x= 2/3+i 1/3 ,x= 2/3-i 1/3

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12 x - 5 = 9 x^{2}

x = \frac{2}{3} + i \frac{1}{3}, x = \frac{2}{3} - i \frac{1}{3}

Solution steps

12 x - 5 = 9 x^{2}

Switch sides

9 x^{2} = 12 x - 5

Move

5

to the left side

9 x^{2} + 5 = 12 x

Move

12 x

to the left side

9 x^{2} + 5 - 12 x = 0

Write in the standard form

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

9 x^{2} - 12 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 9 \cdot 5}{2 \cdot 9}

Simplify

(- 12)^{2} - 4 \cdot 9 \cdot 5 : 6 i

x_{1, 2} = \frac{- ( - 12 ) \pm 6 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 6 i}{2 \cdot 9}, x_{2} = \frac{- ( - 12 ) - 6 i}{2 \cdot 9}

x = \frac{- ( - 12 ) + 6 i}{2 \cdot 9} : \frac{2}{3} + i \frac{1}{3}

x = \frac{- ( - 12 ) - 6 i}{2 \cdot 9} : \frac{2}{3} - i \frac{1}{3}

The solutions to the quadratic equation are:

x = \frac{2}{3} + i \frac{1}{3}, x = \frac{2}{3} - i \frac{1}{3}

6 x + 29 + 3 x + 124 = 180

\frac{3}{5 i}

5 z^{2} - 24 z - 5

33 (\frac{5}{2} - 1)

(- i)^{3}

What are the solutions to the equation 12x-5=9x^2 ?
The solutions to the equation 12x-5=9x^2 are x= 2/3+i 1/3 ,x= 2/3-i 1/3
Find the zeros of 12x-5=9x^2
The zeros of 12x-5=9x^2 are x= 2/3+i 1/3 ,x= 2/3-i 1/3