How do you solve #x^2-3x-12=0#?

1 Answer
Apr 24, 2017

#x~~5.275#
#x~~-2.275#

Explanation:

There are many ways to solve a quadratic equation.
Look here if you are lost on how to solve quadratic equations.

When in doubt you can't go wrong with using the quadratic formula.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

where #ax^2+bx+c=0#


For #x^2-3x-12=0#

#a=1#
#b=-3#
#c=-12#

Now just substitute into the quadratic formula

#x=(-(-3)+-sqrt((-3)^2-4(1)(-12)))/(2(1))#

#x=(3+-sqrt(9+48))/(2)#

#x=(3+-sqrt(57))/(2)#

Therefore,

#x=(3+sqrt(57))/(2)# #---># #~~5.275#

OR

#x=(3-sqrt(57))/(2)# #---># #~~-2.275#