How do you solve #x^2+4x=9# using the quadratic formula?

2 Answers
Mar 13, 2018

Answer:

#x=(-b+-sqrt(b^2-4ac))/(2a)#
is the quadratic formula where:
a=number in front of x^2 (which is 1)
b= number in front of x (a coefficient, which is 4
c= the constant (which is -9)

Explanation:

First, you want to make sure the equation is in the form;
#ax^2 + bx +c=0# . Do this by moving the 9 to the left side
you then get the equation:
#x^2+4x-9=0#
As stated above, you know the values of a, b and c.
Simply sub these into the quadratic formula.
#x=(-4+-sqrt((4)^2-4xx1xx(-9)))/(2xx1)#

hope it helps

Mar 13, 2018

Answer:

#1.606, -5.606#

Explanation:

Rearrange the formula to equal zero
#x^2+4x-9=0#

The coefficients of the expression are:
a=1, b=4, c=-9

Substitute into the quadratic equation.
#x=(-b+- sqrt(b^2-4ac))/(2a)#

#x=(-4+- sqrt(4^2-4*1*-9))/(2*1)#

#x=(-4+- sqrt(52))/2#

#x=(-4+7.2111)/2# and #(-4-7.2111)/2#