What are the solutions to the equation (5y+2)(y-3)=-13(2+y) ?

The solutions to the equation (5y+2)(y-3)=-13(2+y) are y=2i,y=-2i

Find the zeros of (5y+2)(y-3)=-13(2+y)

The zeros of (5y+2)(y-3)=-13(2+y) are y=2i,y=-2i

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(5 y + 2) (y - 3) = - 13 (2 + y)

y = 2 i, y = - 2 i

Solution steps

(5 y + 2) (y - 3) = - 13 (2 + y)

Expand

(5 y + 2) (y - 3) : 5 y^{2} - 13 y - 6

Expand

- 13 (2 + y) : - 26 - 13 y

5 y^{2} - 13 y - 6 = - 26 - 13 y

Move

6

to the right side

5 y^{2} - 13 y = - 13 y - 20

Move

13 y

to the left side

5 y^{2} = - 20

Divide both sides by

5

y^{2} = - 4

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = - 4, y = - - 4

- 4 = 2 i

- - 4 = - 2 i

y = 2 i, y = - 2 i

x^{3} y^{3}

∣ x - 3 ∣ + 6 \leq 11

\frac{7 ^{6}}{7}

\frac{2}{2} x - \frac{3}{4} \geq \frac{4}{5} - 7 x

x^{2} + 3 > 0

What are the solutions to the equation (5y+2)(y-3)=-13(2+y) ?
The solutions to the equation (5y+2)(y-3)=-13(2+y) are y=2i,y=-2i
Find the zeros of (5y+2)(y-3)=-13(2+y)
The zeros of (5y+2)(y-3)=-13(2+y) are y=2i,y=-2i