Question 339749
let altitude be x
base = 3x+2..
Area of triangle = {{{(1/2) * base*altitude}}}
28 = {{{(1/2)*x*(3x+2)}}}
multiply by 2
2*28= {{{x*(3x+2)}}}
56=3x^2+2x
add -56 to both sides
56-56=3x^2+2x-56
3x^2+2x-56=0
solve using quadratic formula
{{{x=(-b+-sqrt(b^2-4*a*c))/(2*a)}}}
a=3, b=2,c =-56
{{{x1=(-2+sqrt(2^2-4*3*(-56)))/(6)}}}
x1=4 inches
{{{x1=(-2-sqrt(2^2-4*3*(-56)))/(6)}}}

x2=-4.66
ignore negative value.
Altitude = 5
base = 3x+2 = 15+2 =17 inches