2. Example 1. Solve 2x2 + 4x – 16 = 0
Steps Solution
1. Isolate the constant term to the
right side of the equation.
Write in the form of ax2 + bx = c
2x2 + 4x = 16
2. Divide each term by the coefficient
of x2.
The coefficient of x2 is 2.
2x2
2
+
4𝑥
2
=
16
2
x2 + 2x = 8
3. Steps Solution
3. Get the square of the half of x and add
both sides of the equation.
x2 + 2x = 8
2
2
= 1 ; (1)2 = 1
x2 + 2x + 1 = 8 + 1
x2 + 2x + 1 = 9
(x + 1)2 = 9
4. Apply the Square Root Property.
(Square both sides)
x + 1)2 = 9
x + 1 = ±3
5. Solve for the roots. x + 1 = ±3
x + 1 = 3 and x + 1 = -3
x = 3 - 1 and x = -3 - 1
x = 2 and x = -4
4. Example 2. Solve for the roots of 2x2 – 8x = 120
Steps Solution
1. Isolate the constant term to the
right side of the equation.
Write in the form of ax2 + bx = c
2x2 – 8x = 120
2. Divide each term by the coefficient
of x2.
The coefficient of x2 is 2.
2x2
2
-
8𝑥
2
=
120
2
x2 – 4x = 60
5. Steps Solution
3. Get the square of the half of x and add
both sides of the equation.
x2 – 4x = 60
=
−4
2
= -2
= (-2)2 = 4
x2 – 4x + 4 = 60 + 4
x2 – 4x + 4 = 64
(x – 2)2 = 64
4. Apply the Square Root Property.
(Square both sides)
(x – 2)2 = 64
x - 2 = ±8
5. Solve for the roots. x – 2 = 8 and x – 2 = -8
x = 8 + 2 and x = -8 + 2
x = 10 and x = -6
6. Example 3. Find the solution set of 3x2 + 48 = -30x.
Steps Solution
1. Isolate the constant term to the
right side of the equation.
Write in the form of ax2 + bx = c
3x2 + 30x = -48
2. Divide each term by the coefficient
of x2.
The coefficient of x2 is 3.
3x2
3
+
30𝑥
3
=
−48
3
x2 + 10x = -16
7. Steps Solution
3. Get the square of the half of x and add
both sides of the equation.
x2 + 10x = -16
10
2
= 5
(5)2 = 25
x2 + 10x + 25 = -16 + 25
(x + 5)2 = 9
4. Apply the Square Root Property.
(Square both sides)
(x + 5)2 = 9
x + 5 = ±3
5. Solve for the roots. x + 5 = 3 and x + 5 = -3
x = 3 – 5 and x = -3 – 5
x = -2 and x = -8