Answer
{$-\frac{5}{2},2$}
Work Step by Step
Step 1: $2x^{2}+x=10$
Step 2: Subtracting 10 from both sides, $2x^{2}+x-10=10-10$
Step 3: The equation therefore becomes $2x^{2}+x-10=0$
Step 4: Comparing $2x^{2}+x-10=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=2$, $b=1$ and $c=-10$
Step 5: The quadratic formula is:
$x=\frac{-b \pm \sqrt (b^{2}-4ac)}{2a}$
Step 6: Substituting the values of a,b and c in the formula:
$x=\frac{-(1) \pm \sqrt ((1)^{2}-4(2)(-10))}{2(2)}$
Step 7: $x=\frac{-1 \pm \sqrt (1+80)}{4}$
Step 8: $x=\frac{-1 \pm \sqrt (81)}{4}$
Step 9: $x=\frac{-1 \pm 9}{4}$
Step 10: $x=\frac{-1+9}{4}$ or $x=\frac{-1-9}{4}$
Step 11: $x=\frac{8}{4}$ or $x=\frac{-10}{4}$
Step 12: $x=2$ or $x=\frac{-5}{2}$
Step 13: Therefore, the solution set is {$-\frac{5}{2},2$}.
