What are the solutions to the equation (2y+7)(2-y)=0 ?

The solutions to the equation (2y+7)(2-y)=0 are y=-7/2 ,y=2

Find the zeros of (2y+7)(2-y)=0

The zeros of (2y+7)(2-y)=0 are y=-7/2 ,y=2

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(2 y + 7) (2 - y) = 0

y = - \frac{7}{2}, y = 2

Decimal Notation

y = - 3.5, y = 2

Solution steps

(2 y + 7) (2 - y) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 y + 7 = 0 or 2 - y = 0

Solve

2 y + 7 = 0 : y = - \frac{7}{2}

Solve

2 - y = 0 : y = 2

The solutions to the quadratic equation are:

y = - \frac{7}{2}, y = 2

y, 20 x^{3} y + 6 y = 5 x^{4} + 6 x + 3 y^{2}

4 m^{2} - 18 = 3 m

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

- x^{2} - 7 x = - 18

x^{2} + 10 x + \frac{25}{2} = 0

What are the solutions to the equation (2y+7)(2-y)=0 ?
The solutions to the equation (2y+7)(2-y)=0 are y=-7/2 ,y=2
Find the zeros of (2y+7)(2-y)=0
The zeros of (2y+7)(2-y)=0 are y=-7/2 ,y=2