Given:
Find: height and base.
Solution: Use given formula for the area:
[tex]A=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}.[/tex]
Substitute [tex]\text{height}=x+5,[/tex] [tex]\text{base}=3x+9[/tex] and solve the equation:
[tex]120=\dfrac{1}{2}\cdot (3x+9)\cdot(x+5),\\ \\80=(x+3)(x+5),\\ \\x^2+3x+5x+15-80=0,\\ \\x^2+8x-65=0,\\ \\D=8^2-4\cdot (-65)=64+260=324,\\ \\x_1=\dfrac{-8-18}{2}=-13,\ x_2=\dfrac{-8+18}{2}=5[/tex]
Solution [tex]x_1=-13[/tex] is extra, because then height and base have negative lengths.
When [tex]x_2=5,[/tex]