Certainly! Let's solve the quadratic equation [tex]\( 11y^2 - 12y + 4 = 6y^2 \)[/tex] step-by-step.
### Step 1: Move all terms to one side of the equation
First, we want to move all terms to one side so that we have zero on the other side. To do this, we'll subtract [tex]\( 6y^2 \)[/tex] from both sides of the equation:
[tex]\[
11y^2 - 12y + 4 - 6y^2 = 0
\][/tex]
### Step 2: Simplify the equation
Combine like terms:
[tex]\[
(11y^2 - 6y^2) - 12y + 4 = 0
\][/tex]
This simplifies to:
[tex]\[
5y^2 - 12y + 4 = 0
\][/tex]
### Step 3: Identify coefficients for the quadratic formula
The quadratic formula is given by:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
In our equation [tex]\(5y^2 - 12y + 4 = 0\)[/tex], the coefficients are:
- [tex]\(a = 5\)[/tex]
- [tex]\(b = -12\)[/tex]
- [tex]\(c = 4\)[/tex]
### Step 4: Calculate the discriminant
The discriminant ([tex]\(\Delta\)[/tex]) is calculated as:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute the values of the coefficients:
[tex]\[
\Delta = (-12)^2 - 4 \times 5 \times 4
\][/tex]
[tex]\[
\Delta = 144 - 80
\][/tex]
[tex]\[
\Delta = 64
\][/tex]
### Step 5: Apply the quadratic formula
Now we substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex] back into the quadratic formula:
[tex]\[
y = \frac{-(-12) \pm \sqrt{64}}{2 \times 5}
\][/tex]
[tex]\[
y = \frac{12 \pm 8}{10}
\][/tex]
### Step 6: Solve for the two possible solutions
We have two solutions to consider:
1. When we use the positive root:
[tex]\[
y = \frac{12 + 8}{10} = \frac{20}{10} = 2
\][/tex]
2. When we use the negative root:
[tex]\[
y = \frac{12 - 8}{10} = \frac{4}{10} = \frac{2}{5}
\][/tex]
### Conclusion
The two solutions to the quadratic equation [tex]\( 11y^2 - 12y + 4 = 6y^2 \)[/tex] are:
[tex]\[
y = 2 \quad \text{and} \quad y = \frac{2}{5}
\][/tex]
These are the values of [tex]\(y\)[/tex] that satisfy the original equation.
