Certainly! Let's solve the quadratic equation [tex]\(6y^2 + 7y - 20 = 0\)[/tex] step by step.
### Step-by-Step Solution
1. Identify the coefficients:
The given quadratic equation is in the form [tex]\(ay^2 + by + c = 0\)[/tex], where
- [tex]\(a = 6\)[/tex]
- [tex]\(b = 7\)[/tex]
- [tex]\(c = -20\)[/tex]
2. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = 7^2 - 4 \cdot 6 \cdot (-20)
\][/tex]
[tex]\[
\Delta = 49 + 480
\][/tex]
[tex]\[
\Delta = 529
\][/tex]
3. Check the discriminant:
Since [tex]\(\Delta > 0\)[/tex], the quadratic equation has two distinct real solutions.
4. Use the quadratic formula:
The solutions [tex]\(y\)[/tex] of the quadratic equation [tex]\(ay^2 + by + c = 0\)[/tex] are given by:
[tex]\[
y = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
5. Calculate the solutions:
Substituting [tex]\(\Delta = 529\)[/tex], [tex]\(a = 6\)[/tex], and [tex]\(b = 7\)[/tex]:
[tex]\[
y_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-7 + \sqrt{529}}{2 \cdot 6}
\][/tex]
[tex]\[
y_1 = \frac{-7 + 23}{12} = \frac{16}{12} = \frac{4}{3}
\][/tex]
Similarly,
[tex]\[
y_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-7 - \sqrt{529}}{2 \cdot 6}
\][/tex]
[tex]\[
y_2 = \frac{-7 - 23}{12} = \frac{-30}{12} = \frac{-5}{2}
\][/tex]
The solutions to the quadratic equation [tex]\(6y^2 + 7y - 20 = 0\)[/tex] are [tex]\(y_1 = \frac{4}{3}\)[/tex] and [tex]\(y_2 = \frac{-5}{2}\)[/tex].
### Summary
- The discriminant [tex]\(\Delta\)[/tex] is 529.
- The two solutions are [tex]\(\frac{4}{3}\)[/tex] and [tex]\(\frac{-5}{2}\)[/tex].
