What are the solutions to the equation (3y-1)(y+2)=0 ?

The solutions to the equation (3y-1)(y+2)=0 are y= 1/3 ,y=-2

Find the zeros of (3y-1)(y+2)=0

The zeros of (3y-1)(y+2)=0 are y= 1/3 ,y=-2

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(3 y - 1) (y + 2) = 0

y = \frac{1}{3}, y = - 2

Decimal Notation

y = 0.33333 \dots, y = - 2

Solution steps

(3 y - 1) (y + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 y - 1 = 0 or y + 2 = 0

Solve

3 y - 1 = 0 : y = \frac{1}{3}

Solve

y + 2 = 0 : y = - 2

The solutions to the quadratic equation are:

y = \frac{1}{3}, y = - 2

(x + 7)^{2} + 7 (x + 7) + 12 = 0

(4 x + 4) (a x - 1) - x^{2} + 4 = b x

x, x^{2} y^{2} + 3 x y = 1

x^{2} + \frac{3}{2} x = 0

(5 x + 2)^{2} = 32

What are the solutions to the equation (3y-1)(y+2)=0 ?
The solutions to the equation (3y-1)(y+2)=0 are y= 1/3 ,y=-2
Find the zeros of (3y-1)(y+2)=0
The zeros of (3y-1)(y+2)=0 are y= 1/3 ,y=-2