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

2 Answers
Mar 9, 2016

Answer:

The solutions are:

# x =3#

# x = -4#

Explanation:

#x^2 + x-12 =0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:

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

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (1)^2-(4*1*(-12))#

# = 1 +48 =49#

The solutions are found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-1)+-sqrt(49))/(2*1) = (-1+-7)/2#

# x = (-1+7)/2 = 6/2 =3#

# x = (-1-7)/2 = -8/2 = -4#

The solutions are:

# x =3#

# x = -4#

Mar 9, 2016

Answer:

#x=-4,3#

Explanation:

#x^2+x-12=0#

This is in the form of a Quadratic equation
(in form #ax^2+bx+c=0#)

Where

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

Use Quadratic formula

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

Substitute the values

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

#rarrx=(-1+-sqrt(1-(-48)))/(2)#

#rarrx=(-1+-sqrt(1+48))/(2)#

#rarrx=(-1+-sqrt49)/(2)#

#rarrx=(-1+-7)/2#

Now we have two values for #2#

#x=(-1-7)/2,x=(-1+7)/2#

If we solve for it,We get

#rArrx=-4,3#