How do you find the solution to the quadratic equation #x^2-3x=10#?

Question

Eva Abramson · Accepted Answer

#y = x^2 - 3x - 10 # = (x - p)(x - q).
Find p and q knowing product (c = -10) and sum (-b = 3). Compose factor pairs of (-10)--> (-2, 5). This sum is 3 = -b. Then the 2 real roots are: -2 and 5.

Eli Assouline · Answer

undefined To solve the quadratic equation $x^2 - 3x = 10$, follow these steps:

1. Rewrite the equation in the standard form $ax^2 + bx + c = 0$.
2. Subtract 10 from both sides to move all terms to one side of the equation: $x^2 - 3x - 10 = 0$.
3. Identify the values of $a$, $b$, and $c$: $a = 1$, $b = -3$, and $c = -10$.
4. Use the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.
5. Plug in the values of $a$, $b$, and $c$ into the quadratic formula.
6. Solve for $x$ by performing the calculations.
7. The solutions will be the values of $x$ obtained from the quadratic formula.

Eva Adamson · Answer

undefined To find the solution to the quadratic equation $x^2 - 3x = 10$, you can rearrange the equation to set it equal to zero:

$$x^2 - 3x - 10 = 0$$

Then, you can use the quadratic formula, which states that for an equation in the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

For the equation $x^2 - 3x - 10 = 0$, $a = 1$, $b = -3$, and $c = -10$. Plugging these values into the quadratic formula:

$$x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4(1)(-10)}}}}{{2(1)}}$$

$$x = \frac{{3 \pm \sqrt{{9 + 40}}}}{2}$$

$$x = \frac{{3 \pm \sqrt{49}}}{2}$$

$$x = \frac{{3 \pm 7}}{2}$$

So, the solutions to the equation are:

$$x_1 = \frac{{3 + 7}}{2} = \frac{{10}}{2} = 5$$
$$x_2 = \frac{{3 - 7}}{2} = \frac{{-4}}{2} = -2$$

Therefore, the solutions to the quadratic equation $x^2 - 3x = 10$ are $x = 5$ and $x = -2$.