What are the solutions to the equation 9x^2-3x=-4 ?

The solutions to the equation 9x^2-3x=-4 are x= 1/6+i(sqrt(15))/6 ,x= 1/6-i(sqrt(15))/6

Find the zeros of 9x^2-3x=-4

The zeros of 9x^2-3x=-4 are x= 1/6+i(sqrt(15))/6 ,x= 1/6-i(sqrt(15))/6

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9 x^{2} - 3 x = - 4

x = \frac{1}{6} + i \frac{15}{6}, x = \frac{1}{6} - i \frac{15}{6}

Solution steps

9 x^{2} - 3 x = - 4

Move

4

to the left side

9 x^{2} - 3 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

Simplify

(- 3)^{2} - 4 \cdot 9 \cdot 4 : 315 i

x_{1, 2} = \frac{- ( - 3 ) \pm 3 15 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 3 15 i}{2 \cdot 9}, x_{2} = \frac{- ( - 3 ) - 3 15 i}{2 \cdot 9}

x = \frac{- ( - 3 ) + 3 15 i}{2 \cdot 9} : \frac{1}{6} + i \frac{15}{6}

x = \frac{- ( - 3 ) - 3 15 i}{2 \cdot 9} : \frac{1}{6} - i \frac{15}{6}

The solutions to the quadratic equation are:

x = \frac{1}{6} + i \frac{15}{6}, x = \frac{1}{6} - i \frac{15}{6}

(x + 8)^{2} + 10 = - 50

5 x (3 x + 2) = 5

28 x^{2} - 4 x = 0

y^{2} - 5 = 8 y

x, 3 x^{2} + 2 y^{2} = c

What are the solutions to the equation 9x^2-3x=-4 ?
The solutions to the equation 9x^2-3x=-4 are x= 1/6+i(sqrt(15))/6 ,x= 1/6-i(sqrt(15))/6
Find the zeros of 9x^2-3x=-4
The zeros of 9x^2-3x=-4 are x= 1/6+i(sqrt(15))/6 ,x= 1/6-i(sqrt(15))/6