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

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

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

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

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

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

Decimal Notation

y = - 2, y = 2.5

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

y + 2 = 0 : y = - 2

Solve

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

The solutions to the quadratic equation are:

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

8 x^{2} = 9

- 5 x^{2} + 2 x + 4 = 0

x, y^{3} = c (y^{2} - x^{2})

\frac{x ^{2}}{1 0 ^{2}} + \frac{( 4.8 ) ^{2}}{6 ^{2}} = 1

6 x 52 x - 1 = 9 x - 3 x 15

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