A polynomial having degree 2 is considered a second-degree equation and it is also called a quadratic equation. The standard form of the second-degree equation is ax²+bx+c=0. Delta in the second-degree equation is used to find the type of roots that polynomial has. If the Delta value for a polynomial is greater than zero (Delta>0) then the polynomial has two real distinct roots. If the Delta value for a polynomial is equal to zero (Delta=0) then the polynomial has only one root. If the Delta value is less than zero (Delta<0) then the polynomial has two imaginary roots.

If the quadratic equation is of form ax²+bx+c=0 then the formula for finding the delta is given below-

Delta = b²-4ac

Let’s look at the few sample problems on finding the Delta in second-degree equations

Sample Questions

Question 1: Find the delta for the second-degree equation x² – 10x + 21 = 0

Solution:

Given 

x²-10x+21=0

a=1, b=-10, c=21

Delta=b²-4ac

=(-10)²-4(1)(21)

=100-84

=16>0

Given equation has two distinct real roots.

Question 2: Find the delta for the second-degree equation x² + 5x – 6 = 0

Solution:

Given

x²+5x-6=0

a=1, b=5, c=-6

Delta=b²-4ac

=(5)²-4(1)(-6)

=25+24

=49>0

Given equation has two distinct real roots.

Question 3: Find the delta for the second-degree equation x² + 4x + 4 = 0

Solution:

Given

x²+4x+4=0

a=1, b=4, c=4

Delta=b²-4ac

=(4)²-4(1)(4)

=16-16

Delta=0

Given equation has only one root.

Question 4: Find the delta for the second-degree equation x² + 2x + 1 = 0

Solution:

Given

x²+2x+1=0

a=1, b=2, c=1

Delta=b²-4ac

=(2)²-4(1)(1)

=4-4

Delta=0

Given equation has only one root.

Question 5: Find the delta for the second-degree equation x² + 4x + 5 = 0

Solution:

Given

x²+4x+5=0

a=1, b=4, c=5

Delta=b²-4ac

=(4)²-4(1)(5)

=16-20

=-4<0

Given equation has two imaginary roots.

Question 6: Find the delta for the second-degree equation x² – 2x + 2 = 0

Solution:

Given

x²-2x+2=0

a=1, b=-2, c=2

Delta=b²-4ac

=(2)²-4(1)(2)

=4-8

=-4<0

Given equation has two imaginary roots.