What are the solutions to the equation 3y^2+4y=8y-7 ?

The solutions to the equation 3y^2+4y=8y-7 are y= 2/3+i(sqrt(17))/3 ,y= 2/3-i(sqrt(17))/3

Find the zeros of 3y^2+4y=8y-7

The zeros of 3y^2+4y=8y-7 are y= 2/3+i(sqrt(17))/3 ,y= 2/3-i(sqrt(17))/3

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3 y^{2} + 4 y = 8 y - 7

y = \frac{2}{3} + i \frac{17}{3}, y = \frac{2}{3} - i \frac{17}{3}

Solution steps

3 y^{2} + 4 y = 8 y - 7

Move

7

to the left side

3 y^{2} + 4 y + 7 = 8 y

Move

8 y

to the left side

3 y^{2} - 4 y + 7 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 3 \cdot 7}{2 \cdot 3}

Simplify

(- 4)^{2} - 4 \cdot 3 \cdot 7 : 217 i

y_{1, 2} = \frac{- ( - 4 ) \pm 2 17 i}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 2 17 i}{2 \cdot 3}, y_{2} = \frac{- ( - 4 ) - 2 17 i}{2 \cdot 3}

y = \frac{- ( - 4 ) + 2 17 i}{2 \cdot 3} : \frac{2}{3} + i \frac{17}{3}

y = \frac{- ( - 4 ) - 2 17 i}{2 \cdot 3} : \frac{2}{3} - i \frac{17}{3}

The solutions to the quadratic equation are:

y = \frac{2}{3} + i \frac{17}{3}, y = \frac{2}{3} - i \frac{17}{3}

e^{l n (1)}

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

x^{2} - x - 4 = 2

3.1 x - 1.4 \leq 1.3 x + 6.7

x^{2} - 64 y^{2}

What are the solutions to the equation 3y^2+4y=8y-7 ?
The solutions to the equation 3y^2+4y=8y-7 are y= 2/3+i(sqrt(17))/3 ,y= 2/3-i(sqrt(17))/3
Find the zeros of 3y^2+4y=8y-7
The zeros of 3y^2+4y=8y-7 are y= 2/3+i(sqrt(17))/3 ,y= 2/3-i(sqrt(17))/3