Question 298071


From {{{q^2+4q+3}}} we can see that {{{a=1}}}, {{{b=4}}}, and {{{c=3}}}



{{{D=b^2-4ac}}} Start with the discriminant formula.



{{{D=(4)^2-4(1)(3)}}} Plug in {{{a=1}}}, {{{b=4}}}, and {{{c=3}}}



{{{D=16-4(1)(3)}}} Square {{{4}}} to get {{{16}}}



{{{D=16-12}}} Multiply {{{4(1)(3)}}} to get {{{(4)(3)=12}}}



{{{D=4}}} Subtract {{{12}}} from {{{16}}} to get {{{4}}}



Since the discriminant is greater than zero, this means that there are two real solutions.