Question 281671

From {{{x^2+10x+19}}} we can see that {{{a=1}}}, {{{b=10}}}, and {{{c=19}}}



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



{{{D=(10)^2-4(1)(19)}}} Plug in {{{a=1}}}, {{{b=10}}}, and {{{c=19}}}



{{{D=100-4(1)(19)}}} Square {{{10}}} to get {{{100}}}



{{{D=100-76}}} Multiply {{{4(1)(19)}}} to get {{{(4)(19)=76}}}



{{{D=24}}} Subtract {{{76}}} from {{{100}}} to get {{{24}}}



Since the discriminant is greater than zero, and isn't a perfect square, this means that there are two real and irrational solutions.