y^2-4y-12=0

Please help. :)

The simplest solution is by factorization:

y^2-4y-12=0
(y 2)(y 6)=0
Since 2*6=12, 6-2=4
are what you are looking for.
Now we need to figure out the signs:
-12 means that exactly one of the constants (2 or 6) carry a negative sign.
If we give assign 2 and -6, we get 2-6=-4, which is what we want, so:
(y+2)(y-6)=0
Can you do take it from here?

In google type:

"quadratic equation online"

When you see list of results click on:

webgraphingcom/quadraticequation_quadraticformula.jsp

In rectacangle type: y^2-4y-12=0

and click option solve it

To solve the given quadratic equation, y^2 - 4y - 12 = 0, you can use the quadratic formula or factorization method. Let's go through both methods:

Method 1: Quadratic Formula
The quadratic formula is given by:

y = (-b ± √(b^2 - 4ac))/(2a)

For the given equation, the coefficients are:
a = 1
b = -4
c = -12

Substituting these values into the quadratic formula, we get:

y = (-(-4) ± √((-4)^2 - 4(1)(-12)))/(2(1))
= (4 ± √(16 + 48))/2
= (4 ± √64)/2
= (4 ± 8)/2

There are two possible solutions:

1. When y = (4 + 8)/2
Simplifying, we get:
y = 12/2
y = 6

2. When y = (4 - 8)/2
Simplifying, we get:
y = -4/2
y = -2

Therefore, the solutions to the quadratic equation y^2 - 4y - 12 = 0 are y = 6 and y = -2.

Method 2: Factorization
To solve the given quadratic equation by factorization, we need to find two numbers whose product is equal to -12 and sum is equal to -4 (coefficient of y).

-4y = 2y - 6y (Factors of 2 that add up to -6 are -2 and -6)
= (2y - 6)(y + 2)

Setting each factor to zero, we get:

2y - 6 = 0 or y + 2 = 0

Solving each equation, we obtain:

2y = 6 y = -2

y = 3

Therefore, the solutions to the quadratic equation y^2 - 4y - 12 = 0 are y = 6, y = -2, and y = 3.

Both methods should give you the correct solutions to the quadratic equation.