Sure! Let's solve the equation [tex]\(2y^2 = 3y + 2\)[/tex] step-by-step.
1. Set the equation to zero:
Starting with [tex]\(2y^2 = 3y + 2\)[/tex], we bring all terms to one side of the equation:
[tex]\[
2y^2 - 3y - 2 = 0
\][/tex]
2. Identify the coefficients:
The standard form of a quadratic equation is [tex]\(ay^2 + by + c = 0\)[/tex]. Here, [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -2\)[/tex].
3. Use the quadratic formula:
The quadratic formula is given by:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substitute [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -2\)[/tex] into the formula:
[tex]\[
y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)}
\][/tex]
4. Simplify under the square root:
[tex]\[
y = \frac{3 \pm \sqrt{9 + 16}}{4}
\][/tex]
[tex]\[
y = \frac{3 \pm \sqrt{25}}{4}
\][/tex]
5. Simplify the square root:
[tex]\[
y = \frac{3 \pm 5}{4}
\][/tex]
6. Solve for the two possible values of y:
- When using [tex]\(+\)[/tex]:
[tex]\[
y = \frac{3 + 5}{4} = \frac{8}{4} = 2
\][/tex]
- When using [tex]\(-\)[/tex]:
[tex]\[
y = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac{1}{2}
\][/tex]
These give us the solutions:
[tex]\[
y = -\frac{1}{2}, 2
\][/tex]
