2=x^2+5x
We must write the equation in the form:
ax^2+bx+c=0
With all the terms in one side of the equation
2-2=x^2+5x-2
0=x^2+5x-2
x^+5x-2=0
Comparing with the general form of a quadratic equation:
Coefficient of x^2: a=1
Coefficient of x: b=5
Independent term: c=-2
Discriminant: D=b^2-4ac
D=5^2-4(1)(-2)
D=25+8
D=33
The discriminant is greater than zero (positive):
D=33>0 (+)
Then the equation has two different real roots
