Question 849226
{{{3x^2-21x+30=0}}} and {{{20y^2+33y+10=0}}} are quadratic equations.
 
Some quadratic equations can be simplified.
Dividing both sides of the equal sign by {{{3}}} we can transform
{{{3x^2-21x+30=0}}} into the simpler equivalent equation
{{{x^2-7x+10=0}}} .
That equation can be easily solved by factoring,
because {{{x^2-7x+10=(x-7)(x-3)}}} ,
so we can rewrite {{{x^2-7x+10=0}}} as the equivalent equation
{{{(x-7)(x-3)=0}}} .
Then, if the product of {{{(x-7)}}} and {{{(x-7)}}} is zero,
either {{{x-7=0}}} --> {{{highlight(x=7)}}}
or {{{x-3=0}}} --> {{{highlight(x=3)}}} .
 
{{{20y^2+33y+10=0}}} cannot be simplified.
Maybe we could solve it by factoring, but it seems difficult.
For such a quadratic equation, I would use the quadratic formula.
The quadratic formula says that the solution to
{{{ax^2+bx+c=0}}} , if it exists, is given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{20y^2+33y+10=0}}} is {{{ax^2+bx+c=0}}} with {{{y}}} instead of {{{x}}} ,
{{{a=20}}} , {{{b=33}}} and {{{c=10}}} so we try to calculate
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} as
{{{y =(-33 +- sqrt(33^2-4*20*10))/(2*20)=(-33 +- sqrt(1089-800))/40=(-33 +- sqrt(289))/40=(-33 +- 17)/40 }}}
The solutions are:
{{{y=(-33+17)/40=-16/40=highlight(-2/5)}}} or {{{y=highlight(-0.4)}}}
and
{{{y=(-33-17)/40=-50/40=highlight(-5/4)}}} or {{{y=highlight(-1.25)}}}