What are the solutions to the equation (4y+3)(2-y)=0 ?

The solutions to the equation (4y+3)(2-y)=0 are y=-3/4 ,y=2

Find the zeros of (4y+3)(2-y)=0

The zeros of (4y+3)(2-y)=0 are y=-3/4 ,y=2

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

y = - \frac{3}{4}, y = 2

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

4 y + 3 = 0 : y = - \frac{3}{4}

Solve

2 - y = 0 : y = 2

The solutions to the quadratic equation are:

y = - \frac{3}{4}, y = 2

4 + 16

7 x + 39 \geq 53

(- 125)^{\frac{4}{3}}

10 x - x^{2} - 16

\frac{1}{2} (1) + 3

What are the solutions to the equation (4y+3)(2-y)=0 ?
The solutions to the equation (4y+3)(2-y)=0 are y=-3/4 ,y=2
Find the zeros of (4y+3)(2-y)=0
The zeros of (4y+3)(2-y)=0 are y=-3/4 ,y=2