1. SOLVING QUADRATIC EQUATIONS
USING THE QUADRATIC FORMULA
BY L.D.

2. TABLE OF CONTENTS
• Slide 3: Formula
• Slide 4: Solve 2x2 – 5x – 3 = 0
• Slide 10: Solve 2x2 + 7x = 9
• Slide 13: Solve x2 + x – 1 = 0

3. FORMULA
The Quadratic Formula:
We start with a problem structured like ax2 + bx + c = 0, then we
use the a, b & c from that format in ournew formula:
x = -b ± √b2 – 4ac
2a

4. PROBLEM 1
• Solve the equation
2x2 – 5x – 3 = 0

5. PROBLEM 1
Formula
• Solve the equation
ax2 + bx + c = 0
2x2 – 5x – 3 = 0 to
The first thing we need to do is to x = -b ± √b – 4ac
2
2a
see if this fits the first formula. It does
since it has the variable squared,
the coefficient and the constant all equaling to zero.

6. PROBLEM 1
Formula
• Solve the equation
ax2 + bx + c = 0
2x2 – 5x – 3 = 0 to
Next we find the a, b & c and put x = -b ± √b – 4ac
2
2a
them into the second formula. To
find them we just match up the
places they are in formula one to the places we
have in our problem. By doing this we discover that
a=2 b = -5 c = -3

7. PROBLEM 1
• Solve the equation Formula
ax2 + bx + c = 0
2x2 – 5x – 3 = 0
to
a=2 b = -5 c = -3 x = -b ± √b2 – 4ac
2a
Now that we have these, we can
place them in our second formula.
x = -(-5) ± √(-5)2 – 4(2)(-3)
2(2)

8. PROBLEM 1
Now we solve the equation we
made!
x=5+7
x = -(-5) ± √(-5)2 – 4(2)(-3) 4
2(2)
x = 12
x = 5± √25 + 24 or x = 3
4
4
x = 5 ± √49 x=5-7
4
4
x = -2
x=5±7
4 or x = -1/2
4

9. PROBLEM 2
• Solve 2x2 + 7x = 9

10. PROBLEM 2
Formula
• Solve 2x2 + 7x = 9
ax2 + bx + c = 0
So first we reorder it!
to
2x2 + 7x = 9 x = -b ± √b2 – 4ac
2a
-9 -9
2x2 + 7x – 9 = 0

11. PROBLEM 2
2x2 + 7x – 9 = 0 Formula
ax2 + bx + c = 0
Now we identify the a, b & c and
to
place it in formula two. Then we
solve. x = -b ± √b2 – 4ac
2a
x = -(7) ± √(7)2 – 4(2)(-9)
2(2)
x = -7 ± √49 – (- 72)
x = -7 ± 11
4 x = -7 ± √121
4
4

12. PROBLEM 2
x = -7 ± 11
4
x = -7 + 11
x = -7 - 11 4
4
x= 4
x = -18 or x = 1
or x = -9/2 4
4

13. PROBLEM 3
• Solve x2 + x – 1 = 0

14. PROBLEM 3
Formula
• Solve x2 + x – 1 = 0
ax2 + bx + c = 0
First we need to identify a, b & c to
x = -b ± √b2 – 4ac
a=1 b=1 c = -1 2a
Next we need to put them in our problem and solve.
x = -(1) ± √(1)2 – 4(1)(-1)
2(1)

15. PROBLEM 3
x = -(1) ± √(1)2 – 4(1)(-1)
2(1)
x = -1 ± √1+ 4
2
We can go no= -1 ± √5
x
further since 5 can’t 2
be squared.
x = -1 + √5 x = -1 - √5
2 2