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

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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)) :}#