how do you solve this problem? below?

Question

the height (h) of a traingle is two less than twice its base(b). the area (as given by the formula.(a=1/2 bh)is 90. what must be the height?

Answer 1

h = 2b - 2
1/2bh = 90 ..........*2
bh = 180
from equ.1
h = 2b - 2
bh = 180
b (2b - 2) = 180
2b² - 2b - 180 = 0
b² - b - 90 = 0
(b - 10)(b + 9) = 0
b = 10 or b = -9 (refused) -as no legnth is -ve -
b = 10
h = 2b - 2
h = 20 - 2 = 18
h = 18 legnth unit & b = 10 legnth unit

Answer 2

let b= the base

Then h= 2b-2
So, b(2b-2)/2 = 90
2b^2 -2b = 180
b^2-2b -90 =0
(b-10)(b+9)=0
b=10=base
2b-2 = 2(10)-2=18 =height

Answer 3

Set up an equation:

You know that: h=2b-2

So plug it in:
A = (1/2)(2b-2)(b) = 90

I'll give you the easy part:

180=(b-1)(b)
0=b^2-b-180

Now just solve for b, then use that to solve for h.

Answer 4

height=2b-2
base=b
1/2(b*(2b-2))=90
Use Distributive property a(b+c)=ab+ac
1/2(2b^2-2b)=90
Use Distributive property(again)
b^2-b=90
Convert to quadradic equation.
b^2-b-90=0
a=1
b=-1
c=-90
x=(b)
Formula: x= (-b±sqrt(b^2-4ac))/2a
Solving, we get 0.5±9.5, so b is either 9 or 10 and the height is either 16 or 18. However, 1/2(9)(16) equals 72, so, b=10 and h=18.

Answer 5

h=2-2b=2(1-b)
A=90=bh/2-->180=bh-->b=180/h

h=2(1-180/h)
h=2-360/h
h^2=2h-360
h^2-2h+360=0
(h-20)(h+18)=0

h=20 or
h=18