Question 132299
{{{y^2=(5/7)y+3/7}}} Start with the given equation



{{{0=(5/7)y+3/7-y^2}}} Subtract {{{y^2}}} from both sides



{{{0=-y^2+(5/7)y+3/7}}} Rearrange the terms



{{{7*0=7(-y^2+(5/7)y+3/7)}}} Multiply both sides by the LCD 7. This will eliminate the fractions



{{{0=-7y^2+5y+3}}} Distribute and multiply



From the quadratic formula


{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


the discriminant consists of all of the terms in the square root. So the discriminant is


{{{D=b^2-4ac}}}


the discriminant tells us how many solutions (and what type of solutions) we can expect for any quadratic.



Now let's find the discriminant for {{{y=-7y^2+5y+3}}}:


{{{D=b^2-4ac}}} Start with the given equation


{{{D=(5)^2-4*-7*3}}} Plug in a=-7, b=5, c=3


{{{D=25-4*-7*3}}} Square 5 to get 25


{{{D=25+84}}} Multiply -4*-7*3 to get 84


{{{D=109}}} Combine 25 and 84 to get 109



Since the discriminant equals 109  (which is greater than zero) , this means  there are two real solutions. Remember if the discriminant is greater than zero, then the quadratic will have two real solutions.