how to solve these three problems using the quadratic formula

Question

x²+x-12=0

 

x²+3x-5=0

 

x²+4x+7=0

Answer 1

x = -b ± √b² - 4ac

             2a

 

a = number in front of x²

b = number in front of x

c = number in last term

 

x²+x-12=0

 

x = -(1) ± √(1)² - 4(1)(-12)

                  2(1)

x = -1 ± √1 + 48     <---- two negatives make a positive

              2

x = -1 ± √49
              2

x = -1 ± 7
          2

x = -1 + 7 = 6 = 3 OR x = -1 - 7 = -8 = -4
          2        2                      2        2

Plug 3 and -4 into the equation to figure out which one works:

(3)²+(3)-12=0

9 + 3 - 12 = 0 

12 - 12 = 0  so, 3 works

 

(-4)² + (-4) - 12 = 0

16 - 4 - 12 = 0

12 - 12 = 0 so, -4 also works.... thus x = 3 OR -4

 

x²+3x-5=0

 

x = -(3) ± √(3)² - 4(1)(-5) 

                  2(1)

x = -3 ± √9 + 20 
                2

x = -3 ± √29 
            2

x = -3 + 5.385 = 2.385 = 1.193 OR x = - 3 - 5.385 = -8.385 = -4.193
            2              2                                    2              2

Both work for the most part besides some decimals, so x = 1.193 OR -4.193

 

x²+4x+7=0

 

x = -(4) ± √(4)² - 4(1)(7)  = -4 ± √16 - 28 = -4 ± √-12  

                  2(1)                            2                     2

x = unsolvable because you cannot take the square root of a negative number