What are the solutions to the equation (2y+5)(y-4)=0 ?

The solutions to the equation (2y+5)(y-4)=0 are y=-5/2 ,y=4

Find the zeros of (2y+5)(y-4)=0

The zeros of (2y+5)(y-4)=0 are y=-5/2 ,y=4

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(2 y + 5) (y - 4) = 0

y = - \frac{5}{2}, y = 4

Solution steps

(2 y + 5) (y - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 y + 5 = 0 or y - 4 = 0

Solve

2 y + 5 = 0 : y = - \frac{5}{2}

Solve

y - 4 = 0 : y = 4

The solutions to the quadratic equation are:

y = - \frac{5}{2}, y = 4

x^{2} = 21 x

(7 x - 2)^{2} = 49

22 = - 2 u

3 x^{2} - 32 x + 64

x^{4} - 8 x^{2} + 16

What are the solutions to the equation (2y+5)(y-4)=0 ?
The solutions to the equation (2y+5)(y-4)=0 are y=-5/2 ,y=4
Find the zeros of (2y+5)(y-4)=0
The zeros of (2y+5)(y-4)=0 are y=-5/2 ,y=4