Question 318628


From {{{10x^2+5x-5}}} we can see that {{{a=10}}}, {{{b=5}}}, and {{{c=-5}}}



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



{{{D=(5)^2-4(10)(-5)}}} Plug in {{{a=10}}}, {{{b=5}}}, and {{{c=-5}}}



{{{D=25-4(10)(-5)}}} Square {{{5}}} to get {{{25}}}



{{{D=25--200}}} Multiply {{{4(10)(-5)}}} to get {{{(40)(-5)=-200}}}



{{{D=25+200}}} Rewrite {{{D=25--200}}} as {{{D=25+200}}}



{{{D=225}}} Add {{{25}}} to {{{200}}} to get {{{225}}}



So the discriminant is {{{D=225}}}



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