Question 809383
Not all quadratic equations can be solved by factoring.
Factoring works when the solutions are rational numbers.
That is not the case for {{{4q^2+12q-9=0}}} .
I would solve that by completing the square,
or by applying the quadratic formula.
 
Completing the square:
{{{4q^2+12q-9=0}}} --> {{{4q^2+12q=9}}}
Looking at {{{4q^2+12q}}} , you realize that it is part of
{{{2q+3)^2=4q^2+12q+9}}} ,
so you add 9 to both sides of {{{4q^2+12q=9}}} to get
{{{4q^2+12q+9=9+9}}} --> {{{2q+3)^2=18}}}
Either
{{{2q+3=sqrt(18)}}}-->{{{2q+3=3sqrt(2)}}}-->{{{2q=-3+3sqrt(2)}}}-->{{{highlight(q=(-3+3sqrt(2))/2)}}}
or
{{{2q+3=-sqrt(18)}}}-->{{{2q+3=-3sqrt(2)}}}-->{{{2q=-3-3sqrt(2)}}}-->{{{highlight(q=(-3-3sqrt(2))/2)}}}
Both solutions can bew expressed with the formula
{{{highlight(q=(-3 +- 3sqrt(2))/2)}}}
 
Applying the quadratic formula works well too, but it is kind of boring.
(I only use it when the square to be completed is not too obvious).
Assuming that you remember the formula and apply it flawlessly, you end up with
{{{q=(-12 +- sqrt(288))/8=(-3*4 +- sqrt(144*2))/(2*4)=(-3*4 +- sqrt(144)*sqrt(2))/(2*4)=(-3*4 +- 12*sqrt(2))/(2*4)=(-3*4 +- 3*4*sqrt(2))/(2*4)=
4*(-3 +- 3*sqrt(2))/(2*4)=highlight((-3 +- 3sqrt(2))/2)}}}