How do i solve quadratics and quadratic formula

Step-by-step explanation:

Multiply the first and last coefficient

Find a pair that has the product of ac (above) and the sum of b

Replace the b-term with the pair.

Divide the four terms into two-terms on the left side and two terms on the right side and factor out the common terms from each side.

This is reverse of the distributive property so the terms factored out (on the outside of the common term) combine to create a factor and the common term is the other factor.  

Set each factor equal to zero and solve for the variable.

1a. a = 1/2      a = -1

2a² + 3a + 1 = 0      → ac=2, b = 3    → use 1, 2  (product is 2 & sum is 3)

2a² + 2a + 1a + 1 = 0

2a² + 2a     +1a + 1        = 0

2a(a + 1)      +1(a + 1)      = 0

(2a + 1)(a + 1) = 0

2a + 1 = 0       and         a + 1 = 0

       a = 1/2    and               a = -1

1b. y = -5/2    y = -5

2y² + 5y - 25 = 0  → ac=-50, b = 5 → use 10, -5  (product is -50 & sum is -5)

2y² + 10y - 5y - 25 = 0

2y² + 10y      - 5y - 25        = 0

2y(y + 5)      -5(y + 5)      = 0

(2y - 5)(y + 5) = 0

2y + 5 = 0       and         y + 5 = 0

       y = -5/2    and               y = -5

1c. y = 5     y = 1/2

2y² - 11y + 5 = 0  → ac=10, b = -11 → use -1, -10  (product is 10 & sum is -11)

2y² - 1y - 10y + 5 = 0

2y² - 1y       -10y + 5        = 0

y(2y - 1)      -5(2y - 1)      = 0

(y - 5)(2y - 1) = 0

y - 5 = 0       and         2y - 1 = 0

     y = 5       and               y = 1/2

[ 1 + (2/3)(7 - 1)  ,  4 + (2/3)(1 - 4) ]  =

[ 1 + (2/3)(6) , 4 + (2/3)(-3) ]  =

[ 1 + 4.   4  - 2 ]  =

( 5 , 2)

Step-by-step explanation:

:)