Solve the quadratic equation -10x^2 + 5x + 19 = 0.

Step-by-step solution

1. 1

   Read off the coefficients as they stand. With $a = -10$, $b = 5$, $c = 19$, there is no need to multiply through by $-1$ first — the quadratic formula handles a negative leading coefficient directly, as long as the sign is carried into $2a$.

2. 2

   Compute the discriminant.

   $b^2 - 4ac = 25 - 4(-10)(19) = 25 + 760 = 785$

   It is positive, so there are two distinct real roots. The double negative in $-4ac$ is where sign errors usually creep in: $-4 \times (-10) \times 19$ is $+760$, not $-760$.

3. 3

   Check whether the surd simplifies. $785 = 5 \times 157$, and $157$ is prime. Neither factor is repeated, so $\sqrt{785}$ has no perfect-square factor to pull out and stays as it is; numerically $\sqrt{785} \approx 28.017851$.

4. 4

   Apply the formula and tidy the signs.

   $x = \frac{-5 \pm \sqrt{785}}{2(-10)} = \frac{-5 \pm \sqrt{785}}{-20} = \frac{5 \mp \sqrt{785}}{20}$

   Since the $\pm$ covers both signs anyway, this is the same set as $\dfrac{5 \pm \sqrt{785}}{20}$.

5. 5

   Evaluate and verify with Vieta.

   $x_1 = \frac{5 + 28.017851}{20} \approx 1.650893, \qquad x_2 = \frac{5 - 28.017851}{20} \approx -1.150893$

   Their sum is $0.500000 = -\tfrac{b}{a} = \tfrac{5}{10}$ and their product is $-1.900000 = \tfrac{c}{a} = \tfrac{19}{-10}$. Substituting $x_1$ into the quadratic returns $0$ to within $4 \times 10^{-15}$.