How do you solve #x^2-2x-10=5#?

Question

Julia Adams · Accepted Answer

-3 and 5

#y = x^2 - 2x - 15 = 0# Find 2 real roots, with opposite signs (ac < 0), knowing sum (-b = 2) and product (c = - 15). They are: -3 and 5. (because sum = -b = 2 and product - 15) NOTE. When a = 1 it isn't necessary to use the lengthy solving by grouping and solving the 2 binomials.

Hazel Anglin · Answer

undefined To solve the equation $x^2 - 2x - 10 = 5$:

1. First, move the constant term to the other side to set the equation to zero: $x^2 - 2x - 10 - 5 = 0$, which simplifies to $x^2 - 2x - 15 = 0$.
2. Now, apply the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a = 1$, $b = -2$, and $c = -15$.
3. Substitute the coefficients into the quadratic formula: $x = \frac{{-(-2) \pm \sqrt{{(-2)^2 - 4(1)(-15)}}}}{{2(1)}}$.
4. Simplify inside the square root: $x = \frac{{2 \pm \sqrt{{4 + 60}}}}{2}$, which becomes $x = \frac{{2 \pm \sqrt{{64}}}}{2}$.
5. Further simplify: $x = \frac{{2 \pm 8}}{2}$.
6. There are two possible solutions:
   a) $x = \frac{{2 + 8}}{2} = \frac{{10}}{2} = 5$.
   b) $x = \frac{{2 - 8}}{2} = \frac{{-6}}{2} = -3$.

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