A quadratic function is represented by the following equation:

${\displaystyle y=ax}$2${\displaystyle +bx+c}$
• ${\displaystyle ax}$2 = Quadratic Term
• ${\displaystyle bx}$ = Linear Term
• ${\displaystyle c}$ = Constant Term

1. ${\displaystyle x}$ 2 ${\displaystyle +7x+6}$

• Factor them: We get ${\displaystyle (x+6)(x+1)}$ .
• The linear terms must add to make 7.
• The constant terms needs to multiply to make 6.
• Set them out as problems to solve:
• ${\displaystyle x+6=0}$  → x = -6
• ${\displaystyle x+1=0}$  → x = -1
• Your answers are ${\displaystyle -6}$  and ${\displaystyle -1}$ .

2. ${\displaystyle x}$ 2 ${\displaystyle -9}$ .

• Factor them: We get ${\displaystyle (x-3)(x+3)}$ .
• Set them out as problems to solve:
• ${\displaystyle x+3=0}$  → x = -3
• ${\displaystyle x-3=0}$  → x = 3
• Your answers are ${\displaystyle 3}$  and ${\displaystyle -3}$ .

3. ${\displaystyle 2x}$ 2${\displaystyle +12x+10}$

• Divide all of the terms by the GCF: 2: We get a new problem to deal with, which is ${\displaystyle x}$ 2${\displaystyle +6x+5}$ .
• Factor them: We get ${\displaystyle 2(x+5)(x+1)}$ .
• Set them out as problems to solve:
• ${\displaystyle x+5=0}$  → x = -5
• ${\displaystyle x+1=0}$  → x = -1
• Your answers are ${\displaystyle -5}$  and ${\displaystyle -1}$ .

Solving Quadratic Factors by Completing the Square

1. ${\displaystyle x}$ 2 ${\displaystyle +8x}$  __ = ${\displaystyle 19}$

• Take the Linear Term and divide it by ${\displaystyle 2}$ : We get ${\displaystyle 4}$ .
• We take this number, ${\displaystyle 4}$ , and square it: We get ${\displaystyle 16}$ .
• We add ${\displaystyle 16}$  to ${\displaystyle 19}$ : We get ${\displaystyle 35}$ .
• We now have: ${\displaystyle (x+4)}$ 2 ${\displaystyle =35}$ .
• We square both sides: We get ${\displaystyle x+4=}$ ${\displaystyle 35}$ .
• We minus 4 to the other side. Here is our answer.: ${\displaystyle x=-4}$ ± √${\displaystyle 35}$ .

2. ${\displaystyle x}$ 2 ${\displaystyle +x}$  + ___

• Take the Linear Term and divide it by ${\displaystyle 2}$ : We get ${\displaystyle {\tfrac {1}{2}}}$ .
• We take this number, ${\displaystyle {\tfrac {1}{2}}}$ , and square it: We get ${\displaystyle {\tfrac {1}{4}}}$ .
• We have our answer: ${\displaystyle x}$ 2 ${\displaystyle +x}$  + ${\displaystyle {\tfrac {1}{4}}}$ .

3. ${\displaystyle x}$ 2 ${\displaystyle +45}$  = ${\displaystyle 10x}$

• Rearrange this problem so that it matches the standard format for a quadratic equation: We switch the ${\displaystyle 45}$  and the ${\displaystyle 10x}$  around, forming our new problem: ${\displaystyle x}$ 2 ${\displaystyle -10x}$  = ${\displaystyle -45}$ .
• Divide the Linear Tearm, ${\displaystyle bx}$ , by ${\displaystyle 2}$ : This gives us ${\displaystyle 5}$ .
• Square the ${\displaystyle 5}$ : This gives us ${\displaystyle 25}$ .
• Add the ${\displaystyle 25}$  to ${\displaystyle -45}$ : This brings our problem to (${\displaystyle x}$  ${\displaystyle -5)}$ 2 = ${\displaystyle -20}$ .
• Square both sides of the problem: This brings us to ${\displaystyle x-5=}$ i√${\displaystyle 20}$ .
• Find the square root of ${\displaystyle 20}$  (don't forget the ${\displaystyle i}$ ) and then add ${\displaystyle 5}$  to the opposite side to find your answer: Our final answer is ${\displaystyle x=-5}$ ± ${\displaystyle 2i}$ ${\displaystyle 5}$ .