What are the solutions to the equation (3y-5)(2y+7)=0 ?

The solutions to the equation (3y-5)(2y+7)=0 are y= 5/3 ,y=-7/2

Find the zeros of (3y-5)(2y+7)=0

The zeros of (3y-5)(2y+7)=0 are y= 5/3 ,y=-7/2

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(3 y - 5) (2 y + 7) = 0

y = \frac{5}{3}, y = - \frac{7}{2}

Decimal Notation

y = 1.66666 \dots, y = - 3.5

Solution steps

(3 y - 5) (2 y + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 y - 5 = 0 or 2 y + 7 = 0

Solve

3 y - 5 = 0 : y = \frac{5}{3}

Solve

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

The solutions to the quadratic equation are:

y = \frac{5}{3}, y = - \frac{7}{2}

\frac{x ^{2}}{2} + x - 4 = 0

2 n^{2} - 9 n - 2 = 0

25 t^{2} - 20 t + 4 = 0

x (10 x - 13) = 3

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

What are the solutions to the equation (3y-5)(2y+7)=0 ?
The solutions to the equation (3y-5)(2y+7)=0 are y= 5/3 ,y=-7/2
Find the zeros of (3y-5)(2y+7)=0
The zeros of (3y-5)(2y+7)=0 are y= 5/3 ,y=-7/2