To solve the quadratic equation \(10x^2 + 5x = 5x - 8\), we first rearrange it into standard form \(ax^2 + bx + c = 0\).
Subtract \(5x\) from both sides:
\[ 10x^2 + 5x - 5x = -8 \] \[ 10x^2 = -8 \]
Next, rearranging gives:
\[ 10x^2 + 8 = 0 \]
Now, let's bring -8 to the left side:
\[ 10x^2 + 0x + 8 = 0 \]
We identify the coefficients:
- \(a = 10\)
- \(b = 0\)
- \(c = 8\)
Now, we can use the quadratic formula, which is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values:
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Calculate \(b^2 - 4ac\): \[ b^2 - 4ac = 0^2 - 4(10)(8) = 0 - 320 = -320 \]
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Substitute into the quadratic formula: \[ x = \frac{-0 \pm \sqrt{-320}}{2(10)} = \frac{\pm \sqrt{-320}}{20} \]
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Simplifying \(\sqrt{-320}\): \[ \sqrt{-320} = \sqrt{320} \cdot i = \sqrt{64 \cdot 5} \cdot i = 8\sqrt{5} \cdot i \]
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Thus: \[ x = \frac{\pm 8\sqrt{5} \cdot i}{20} = \frac{8\sqrt{5} \cdot i}{20} = \frac{2\sqrt{5} \cdot i}{5} \]
So, the solutions are:
\[ x = 0 \pm \frac{2\sqrt{5}}{5} i \]
In standard form:
\[ x = 0 + \frac{2\sqrt{5}}{5} i \quad \text{and} \quad x = 0 - \frac{2\sqrt{5}}{5} i \]
Thus, the final solutions are:
\[ x = 0 \pm \frac{2\sqrt{5}}{5} i \]