How do you solve #x^2+4x+3=0# using the quadratic formula?

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Answer 1 · 2017-07-22T12:35:39

#-1" or "-3#

For the quadratic equation

#ax^2+bx+c=0#

the formula for the roots is

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

#(1)" "# Identify the values for #a,b,&" "c#

#x^2+4x+3=0#

cmp

#ax^2+bx+c=0#

#color(red)(a=1)#

#color(blue)(b=4)#

#color(green)(c=3)#

#(2)" "#Substitute these numbers into eh formula

#x=(color(blue)(-4)+-sqrt(color(blue)(4^2)-(4xxcolor(red)(1)xxcolor(green)(3))))/(2xxcolor(red)(1))#

#(3)" "#Carefully proceed and do the calculations

#x=(-4+-sqrt(16-12))/2#

#x=(-4+-sqrt4)/2#

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

now calculate the two separate solutions

#x_1=(-4+2)/2=-2/2=-1#

#x_2=(-4-2)/2=-6/2=-3#