Answer
{$\frac{1 \pm \sqrt (13)}{2}$}
Work Step by Step
Step 1: $0.1x^{2}-0.1x=0.3$
Step 2: Subtracting $0.3$ from both sides of the equation, $0.1x^{2}-0.1x-0.3=0.3-0.3$
Step 3: $0.1x^{2}-0.1x-0.3=0$
Step 4: For simplification, we will multiply both sides of the equation $0.1x^{2}-0.1x-0.3=0$ by 10.
Step 5: $10\times(0.1x^{2}-0.1x-0.3)=10\times0$
Step 6: The equation becomes $x^{2}-x-3=0$
Step 7: Comparing $x^{2}-x-3=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=1$, $b=-1$ and $c=-3$
Step 8: The quadratic formula is:
$x=\frac{-b \pm \sqrt (b^{2}-4ac)}{2a}$
Step 9: Substituting the values of a,b and c in the formula:
$x=\frac{-(-1) \pm \sqrt ((-1)^{2}-4(1)(-3))}{2(1)}$
Step 10: $x=\frac{1 \pm \sqrt (1+12)}{2}$
Step 11: $x=\frac{1 \pm \sqrt (13)}{2}$
Step 12: Therefore, the solution set is {$\frac{1 \pm \sqrt (13)}{2}$}.
