finding the base and height of a triangle

[chabreya]

a triangle has a height that is 3 more than twice the base and the area is 45 square units. Find the base and heights

 

[pka]

\(\displaystyle \L \begin{array}{l}
Area = \frac{{bh}}{2} \\
45 = \frac{{b\left( {2b + 3} \right)}}{2} \\
\end{array}\)

 

[chabreya]

the answer i got is 2.5 =2b^2+3b. I know this is not right what am i doing wrong.

 

[skeeter]

wrong equation ... multiply both sides of pka's equation by 2, you get ...

90 = 2b² + 3b

solve this quadratic equation ... remember that you're looking for a base distance.