Quadratic Equation by Completing the Square

Answer explanation

  $14+x-3x^2=4$    $3x^2-x-14+4=0$  
 $\frac{3x^2}{3}-\frac{x}{3}-\frac{10}{3}=0$ 
 $x^2-\frac{1}{3}x+\left(\frac{1}{6}\right)^2-\left(\frac{1}{6}\right)^2-\frac{10}{3}=0$ 
 $\left(x-\frac{1}{6}\right)^2-\frac{1}{36}-\left(\frac{10}{3}\times\frac{12}{12}\right)$  

 $\left(x-\frac{1}{6}\right)^2-\frac{1}{36}-\frac{120}{36}=0$  

 $\left(x-\frac{1}{6}\right)^2-\frac{121}{36}=0$   $\sqrt{\left(x-\frac{1}{6}\right)^2}=\sqrt{\frac{121}{36}}$    $x-\frac{1}{6}=\pm\frac{11}{6}$   $x=\pm\frac{11}{6}+\frac{1}{6}$   $x=2,\frac{-5}{3}$  

Answer explanation

 $\left(x+1\right)^2=9$  
 $\sqrt{\left(x+1\right)^2}=\sqrt{9}$  
 $x+1=\pm3$  
 $x=3-1=2$ 
 $x=-3-1=-4$  

Answer explanation

 $x^2+12x+\left(\frac{12}{2}\right)^2-\left(\frac{12}{2}\right)^2$   $x^2+12x+36-36$ 
  $\left(x+6\right)^2-36$  

Answer explanation

 $x^2-6x+\left(\frac{-6}{2}\right)^2-\left(\frac{-6}{2}\right)^2+1$   $x^2-6x+9-9+1$ 
  $\left(x-3\right)^2-8$  

Answer explanation

 $x^2+3x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2-2$   $x^2+3x+\frac{9}{4}-\frac{9}{4}-2$ 
  $\left(x+\frac{3}{2}\right)^2-\frac{17}{4}$  

Solve the following quadratics equation by completing the square...

x² - 12x + 31 = 0

• x = -6 ± √5

• x = 6 ± √5

• x = 5 ± √6

• x = -5 ± √6

Solve the following quadratics equation by completing the square... x² + 2x - 5 = 0

• x = -1 ± √6

• x = 1 ± √6

• x = 6 ± √1

• x = -6 ± √1

Complete the Square
x² + 6x = 5

• (x + 3)² = 5

• (x + 6)² = 9

• (x + 3)² = 14

• (x + 6)² = 14

Find the missing value "c" to create a perfect square trinomial:
x² + 26x + ___ 

• 13

• 136

• 169

• 26

What do you add to complete the square?
x² − 12x

• 12

• 6

• 36

• − 6