Question 264673
Use formula {{{(1/2)b(h)=a}}} where b=base h=hieght and a=area 
a=33
{{{(1/2)b(h)=33}}}
b=2h-1
{{{(1/2)(2h-1)(h)=33}}}
{{{(2)(1/2)(2h-1)(h)=33(2)}}}
{{{(2h-1)(h)=66}}}
{{{(2h^2-1h)=66}}}
{{{2h^2-1h-66=0}}} Follow the solver below to get h=(6 or -5)
b=2h-1
{{{b=2(6)-1}}}
{{{b=11}}}
b=2h-1
{{{b=2(-5.5)-1}}}
{{{b=-12}}}
either (b=13)&(h=6) or (b=-12)&(h=-5.5)
.
Checking answers
{{{(1/2)b(h)=33}}}
{{{(1/2)11(6)=33}}}
{{{(1/2)(66)=33}}}
.
{{{(1/2)b(h)=33}}}
{{{(1/2)-12(-5.5)=33}}}
{{{(1/2)(66)=33}}}

*[invoke quadratic "h", 2, -1, -66]