To solve the quadratic equation using the quadratic formula, we first identify the coefficients \(a\), \(b\), and \(c\) from the equation \(ay^2 + by + c = 0\). Then, we apply the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
This will give us the solutions for \(y\).
The given quadratic equation is:
\[
\frac{2}{5} y^{2} + \frac{3}{5} y + \frac{3}{5} = 0
\]
This can be compared to the standard form of a quadratic equation:
\[
ay^2 + by + c = 0
\]
From this, we identify the coefficients as follows:
- \( a = \frac{2}{5} \)
- \( b = \frac{3}{5} \)
- \( c = \frac{3}{5} \)
The quadratic formula is given by:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
The discriminant \(\Delta\) is calculated as:
\[
\Delta = b^2 - 4ac
\]
Substitute the values of \(a\), \(b\), and \(c\):
\[
\Delta = \left(\frac{3}{5}\right)^2 - 4 \times \frac{2}{5} \times \frac{3}{5}
\]
\[
\Delta = \frac{9}{25} - \frac{24}{25}
\]
\[
\Delta = \frac{9 - 24}{25} = \frac{-15}{25} = -\frac{3}{5}
\]
Since the discriminant is negative, the roots will be complex. Substitute the values into the quadratic formula:
\[
y = \frac{-\frac{3}{5} \pm \sqrt{-\frac{3}{5}}}{2 \times \frac{2}{5}}
\]
First, calculate the denominator:
\[
2a = 2 \times \frac{2}{5} = \frac{4}{5}
\]
Now, calculate the roots:
\[
y = \frac{-\frac{3}{5} \pm \sqrt{-\frac{3}{5}}}{\frac{4}{5}}
\]
Simplify the expression:
\[
y = \frac{-\frac{3}{5}}{\frac{4}{5}} \pm \frac{\sqrt{-\frac{3}{5}}}{\frac{4}{5}}
\]
\[
y = -\frac{3}{4} \pm \frac{\sqrt{-\frac{3}{5}}}{\frac{4}{5}}
\]
Since \(\sqrt{-\frac{3}{5}}\) involves an imaginary number, we express it as:
\[
\sqrt{-\frac{3}{5}} = i \sqrt{\frac{3}{5}}
\]
Thus, the roots are:
\[
y = -\frac{3}{4} \pm \frac{i \sqrt{\frac{3}{5}}}{\frac{4}{5}}
\]
Simplify further:
\[
y = -\frac{3}{4} \pm \frac{i \sqrt{3/5} \times 5}{4}
\]
\[
y = -\frac{3}{4} \pm \frac{5i \sqrt{3/5}}{4}
\]
The solutions to the quadratic equation are:
\[
\boxed{y = -\frac{3}{4} \pm \frac{5i \sqrt{3/5}}{4}}
\]
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