What are the solutions to the equation 2y^2+11y=y^2+3y-28 ?

The solutions to the equation 2y^2+11y=y^2+3y-28 are y=-4-2sqrt(3)i,y=-4+2sqrt(3)i

Find the zeros of 2y^2+11y=y^2+3y-28

The zeros of 2y^2+11y=y^2+3y-28 are y=-4-2sqrt(3)i,y=-4+2sqrt(3)i

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2 y^{2} + 11 y = y^{2} + 3 y - 28

y = - 4 - 23 i, y = - 4 + 23 i

Solution steps

2 y^{2} + 11 y = y^{2} + 3 y - 28

Switch sides

y^{2} + 3 y - 28 = 2 y^{2} + 11 y

Move

11 y

to the left side

y^{2} - 8 y - 28 = 2 y^{2}

Move

2 y^{2}

to the left side

- y^{2} - 8 y - 28 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 1 ) ( - 28 )}{2 ( - 1 )}

Simplify

(- 8)^{2} - 4 (- 1) (- 28) : 43 i

y_{1, 2} = \frac{- ( - 8 ) \pm 4 3 i}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 4 3 i}{2 ( - 1 )}, y_{2} = \frac{- ( - 8 ) - 4 3 i}{2 ( - 1 )}

y = \frac{- ( - 8 ) + 4 3 i}{2 ( - 1 )} : - 4 - 23 i

y = \frac{- ( - 8 ) - 4 3 i}{2 ( - 1 )} : - 4 + 23 i

The solutions to the quadratic equation are:

y = - 4 - 23 i, y = - 4 + 23 i

x^{2} + 4 x - 60

\frac{9 X}{8} + 2 = \frac{11 X}{10}

\frac{3}{6} - \frac{1}{3}

8 x^{2} - 18 x - 18

15 x^{3} - 25 x^{2} - 60 x

What are the solutions to the equation 2y^2+11y=y^2+3y-28 ?
The solutions to the equation 2y^2+11y=y^2+3y-28 are y=-4-2sqrt(3)i,y=-4+2sqrt(3)i
Find the zeros of 2y^2+11y=y^2+3y-28
The zeros of 2y^2+11y=y^2+3y-28 are y=-4-2sqrt(3)i,y=-4+2sqrt(3)i