5x^2 + 3x = 2

Let us move 2 to the left side :

==> 5x^2 + 3x -2 =0

Now we have a polynomial in the standard form **ax^2 + bx + c=0**

The discriminant, let us call it (A), gives information weather the polynomial has any roots.

That is, if the discriminant (A)>0 then it has two roots.

if A = 0 , then it has one root.

if A < 0 , then it does not have any REAL roots.

The formula for the discriminant is :

A = b^2 - 4ac

In our case:

a=5 b = 3 c =-2

A = 3^2 - 4*5*-2

= 9 + 40 = 49

**The discriminant (A) = 49 > 0**

**Then the polynomial has 2 real roots.**

To find the discriminant and solve step by step.

Solution:

The discriminant of a genaral quadratic equation ax^2+bx+c is

b^2 - 4ac.

The given equation is 5x^2+3x=2 which could be written as:

5x^2+3x-2 = 0

So a= 5, b=x and c =-2.

Therefore the discriminant of 5x^2+3x-2 is 3^2-4*4*(-2) = 9+40 = 49.

Solution of the equation 5x^2+3x =2 , or 5x^2+3x-2 = 0.

5x^2+**3x**-2 = 0. Group the middle term in the left side to facilitate factor the left.

5x^2+**5x -2x **-2 = 0

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

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

x+1=0, 5x-2 = 0

x = -1 or x = 2/5

An alternate mwthod of solution:

5(x^2+(3/5)x-2/5) = 0

5{[ x^2 + 2(3/(10))+(3/10)^2 ] - (3/10)^2 -2/5 } = 0

5 {(x+3/10)^2 - 49/100} = 0

5{ (x+3/10)^2 - 49/100} = 0. Divide both sides by 5.

(x+3/10)^2 - (7/10)^2 = 0

(x+3/10+7/10)(x+3/10-7/10) = 0, as A^2-B^2= (A+B)(A-B).

(x+1)(x-4/10) = 0

x +1=0 or x-4/10 = 0

x=-1 or x=4/10 =2/5.