What are the solutions to the equation (5y+2)(1+y)=0 ?

The solutions to the equation (5y+2)(1+y)=0 are y=-2/5 ,y=-1

Find the zeros of (5y+2)(1+y)=0

The zeros of (5y+2)(1+y)=0 are y=-2/5 ,y=-1

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(5 y + 2) (1 + y) = 0

y = - \frac{2}{5}, y = - 1

Decimal Notation

y = - 0.4, y = - 1

Solution steps

(5 y + 2) (1 + y) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

5 y + 2 = 0 or 1 + y = 0

Solve

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

Solve

1 + y = 0 : y = - 1

The solutions to the quadratic equation are:

y = - \frac{2}{5}, y = - 1

0 = - 16 t^{2} + 80 t + 4

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

29 (x + 1) + 3 = - x^{2} + 27 x + 6

\frac{350 x ^{2}}{2} = 6.64 \cdot 1257 \cdot (540 - x)

2 (3 h - 1) = 4 - 2 h (h + 5)

What are the solutions to the equation (5y+2)(1+y)=0 ?
The solutions to the equation (5y+2)(1+y)=0 are y=-2/5 ,y=-1
Find the zeros of (5y+2)(1+y)=0
The zeros of (5y+2)(1+y)=0 are y=-2/5 ,y=-1