To solve 2x^2-9x-5=0.

We shall solve this making it as ** X^2 - n^2 =0** or **X^2 = n^2** , where X is is function of x and n is numeric.

2(x^2-4.5x)-5=0

2[x^2-2(4.5/2)x +**(4.5/2)^2**]-** 2(4.5/2)^2**-5=0. Here 2(4.5/2)^2 is added and subtracted with the purpose of making the expression to bring to the form of 2X^2 -n^2 form.

2(x-4.5/2)^2 = 2(4.5/2)^2+5 =15.125

(x-2.25)^2= 15.125/2 =7.5625 which is in X^2=n^2 form.

Take the square root on both sides :

x-2.25 = +2.75 or x-2.25 = -2.75

x=2.25+2.75 = ** 5 ** or

x=2.25-2.75 =** -0.5**

In this solution we will represent x squared as x^2. Therefore the given equation becomes:

2x^2 - 9x - 5 = 0

Therefore: 2x^ - 10x + x - 5 = 0

Therefore: 2x(x - 5) + 1(x - 5) = 0

Therefore:(2x + 1)*(x - 5) = 0

Therefore:(2x + 1) = 0, or (x - 5) = 0

Therefore: x = -1/2 or x = 5 Answer.

2x squared - 9x - 5 = 0

We can solve with the help of quadratic formula, as the question is in quadratic equation form i.e.

2x^2-9x-5=0

Where a=2

b=-9

c=-5

Input the above values into the quadratic formula:

x=[9+sqrt{(-9)^2-4(2)(-5)}]/4 x=[9-sqrt{(-9)^2-4(2)(-5)}]/4

=[9+sqrt(81+40)]/4 =[9-sqrt(81+40)]/4

=[9+sqrt(121)]/4 =[9-sqrt(121)]/4

=(9+11)/4 =(9-11)/4

** =5 =-0.5**

**Solution Set= (5,-0.5)**

`2x^2 - 9x - 5 = 0`

a=2 b=-9 c=-5

`cxxa ` and then factors of `axxc ` that equal b

`2xx-5=-10`

`-10xx1= -10 ` `-10+1=-9`

group:

`(2x^2 - 10x) ( x - 5) `factor out:

`2x(x - 5) + 1(x - 5) = 0 ``(2x + 1) (x - 5)`

set parentheses equal to 0

(2x + 1) = 0

(x - 5) = 0

x = -1/2

x = 5

First of all, let's put the second grade equation under the general form:

a*x^2+b*x+c=0, where a,b,c, are the quotients and x is the variable, or unknown.

Let's identify the quotients of our equation:

a=2, b=-9, c=-5

Now, we cand use the formula for finding the two solution of the equation:

x1=[-b+sqrt(b^2-4*a*c)]/2*a

By substituting the quotients with ours:

x1=[-(-9)+sqrt(81+40)]/2*2

x1=(9+11)/4=20/4=5

x2=[-b-sqrt(b^2-4*a*c)]/2*a

x2=(9-11)/4=-2/4=-1/2