how do I solve (42y^2-7y)(14y^2+53y-45)=0?

Answer 1

(42y^2 - 7y)(14y^2 + 53y - 45)=0

this is true if
42y^2 - 7y = 0
7y(6y - 1) = 0 so x = 0 or x = 1/6
or

14y^2 + 53y -45 = 0
14(y^2 + 53/14 x - 45/14) = 0
14(y^2 + 53/14 x + (53/28)^2 - (53/28)^2 - 45/14) = 0
14((y+53/28)^2 - ( 2809 + 2968)/784) = 0
14 ((y + 53/28)^2 - 5777/784) = 0
14 (y+ 53/28 + sqrt(5777)/28)(y + 53/28 - sqrt(5777)/28) = 0
y = - 53/28 - sqrt(5777)/28
or
y= - 53/28 + sqrt(5777)/28

Answer 2

When you have a polynomial in one variable on one side of an equation, and 0 on the other side, try to factor the polynomial as far as you can (into quadratic (ax^2 + bx + x) or linear (ax + b) expressions), then set each of these factors equal to zero, and solve these separate equations.

So:
(42y^2-7y) factors to 7y * (6y-1), and
(14y^2+53y-45) factors... well, notice that 14 = 2 * 7, and 45 = 9 * 5, or 3 * 15. Considering the - sign, try to find some combination of these that gives 53. 7*9 gets you close, so let's try that first:

7*9 + 2*(-5) = 63 - 10 = 53. Confused? This was just the middle step in multiplying out the factors
(7y-5) * (2y+9) to get (14y^2+53y-45).

The whole thing therefore factors to
7y * (6y-1) * (7y-5) * (2y+9) = 0.

Set each of these four linear expressions equal to 0, and solve each for y to get your four answers.

Didn't have to use the quadratic formula!

Answer 3

Factor it out.

First y*(42y-7)(14y^2+53y-45)=0
so, answers are
y=0, and
42y-7=0 or y = 7/42, and
... I leave the last one for you ;-)

Good luck.

Answer 4

plz try this answer

(-45+53y+14y^2)(-7y+42y^2)


if u satisfied plz tell me