To solve the quadratic equation \(10x^2 + 5x = 5x - 8\), we first need to rearrange it into standard form. Starting from the given equation:
\[ 10x^2 + 5x - 5x + 8 = 0 \] Simplifying this gives:
\[ 10x^2 + 8 = 0 \]
Next, we'll move 8 to the other side:
\[ 10x^2 = -8 \]
Now, divide both sides by 10:
\[ x^2 = -\frac{8}{10} \] \[ x^2 = -\frac{4}{5} \]
Now, we can take the square root of both sides. However, since the right side is negative, we will introduce \(i\) (the imaginary unit):
\[ x = \pm \sqrt{-\frac{4}{5}} = \pm \sqrt{\frac{4}{5}} \cdot i = \pm \frac{2}{\sqrt{5}} i \]
To express the solution in a standard form \(a \pm bi\), we can rationalize the denominator:
\[ x = \pm \frac{2\sqrt{5}}{5} i \]
Thus, the solutions of the equation are:
\[ x = 0 \pm \frac{2\sqrt{5}}{5} i \]
Putting it into the form \(a \pm bi\):
\[ x = 0 \pm \frac{2\sqrt{5}}{5} i \]
Therefore, the solutions in the form \(a \pm bi\) are:
\[ x = 0 + \frac{2\sqrt{5}}{5} i \quad \text{and} \quad x = 0 - \frac{2\sqrt{5}}{5} i \]