How do you solve $x^2-3x=10$ ?

Question

1491 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-05T13:18:22

You could use the quadratic formula.

For a general form quadratic equation

$color(blue)(ax^2 + bx + c = 0)$

you can use the quadratic formula to determine its two solutions

$color(blue)(x_(1,2) = (_b +- sqrt(b^2 - 4ac))/(2a)$

Your quadratic looks like this

$x^2 - 3x = 10$ , which can be rewritten as

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

In your case, you have $a=1$ , $b=-3$ , and $c=-10$ . This means that the two solutions will be

$x_(1,2) = (-(-3) +- sqrt((-3)^2 - 4 * 1 * (-10)))/(2 * 1)$

$x_(1,2) = (3 +- sqrt(49))/2$

$x_(1,2) = (3 +- 7)/2 = {(x_1 = (3+7)/2 = color(green)(5)), (x_2 = (3-7)/2 = color(green)(-2)) :}$

How do you solve x^2-3x=10 x^2-3x=10 ?