What are the solutions to the equation 4y^2-6y+9=0 ?

The solutions to the equation 4y^2-6y+9=0 are y= 3/4+i(3sqrt(3))/4 ,y= 3/4-i(3sqrt(3))/4

Find the zeros of 4y^2-6y+9=0

The zeros of 4y^2-6y+9=0 are y= 3/4+i(3sqrt(3))/4 ,y= 3/4-i(3sqrt(3))/4

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4 y^{2} - 6 y + 9 = 0

y = \frac{3}{4} + i \frac{3 3}{4}, y = \frac{3}{4} - i \frac{3 3}{4}

Solution steps

4 y^{2} - 6 y + 9 = 0

Solve with the quadratic formula

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

Simplify

(- 6)^{2} - 4 \cdot 4 \cdot 9 : 63 i

y_{1, 2} = \frac{- ( - 6 ) \pm 6 3 i}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 6 ) + 6 3 i}{2 \cdot 4}, y_{2} = \frac{- ( - 6 ) - 6 3 i}{2 \cdot 4}

y = \frac{- ( - 6 ) + 6 3 i}{2 \cdot 4} : \frac{3}{4} + i \frac{3 3}{4}

y = \frac{- ( - 6 ) - 6 3 i}{2 \cdot 4} : \frac{3}{4} - i \frac{3 3}{4}

The solutions to the quadratic equation are:

y = \frac{3}{4} + i \frac{3 3}{4}, y = \frac{3}{4} - i \frac{3 3}{4}

y, x = a y^{2} + b y + c

11 x^{2} - 7 x + 1 = 0

\frac{x ^{2}}{2} + 2 x - 6 = 0

x, 3 x^{2} + 4 y^{2} + 3 = 0

16 x^{2} + k x - h = 0

What are the solutions to the equation 4y^2-6y+9=0 ?
The solutions to the equation 4y^2-6y+9=0 are y= 3/4+i(3sqrt(3))/4 ,y= 3/4-i(3sqrt(3))/4
Find the zeros of 4y^2-6y+9=0
The zeros of 4y^2-6y+9=0 are y= 3/4+i(3sqrt(3))/4 ,y= 3/4-i(3sqrt(3))/4