What is the quadratic portion in this quadratic equation 7x2-12x+16=0

x = 6/7 + (2 i sqrt(19))/7 or x = 6/7 - (2 i sqrt(19))/7

Step-by-step explanation:

Solve for x:

7 x^2 - 12 x + 16 = 0

Hint: | Write the quadratic equation in standard form.

Divide both sides by 7:

x^2 - (12 x)/7 + 16/7 = 0

Hint: | Solve the quadratic equation by completing the square.

Subtract 16/7 from both sides:

x^2 - (12 x)/7 = -16/7

Hint: | Take one half of the coefficient of x and square it, then add it to both sides.

Add 36/49 to both sides:

x^2 - (12 x)/7 + 36/49 = -76/49

Hint: | Factor the left hand side.

Write the left hand side as a square:

(x - 6/7)^2 = -76/49

Hint: | Eliminate the exponent on the left hand side.

Take the square root of both sides:

x - 6/7 = (2 i sqrt(19))/7 or x - 6/7 = -(2 i sqrt(19))/7

Hint: | Look at the first equation: Solve for x.

Add 6/7 to both sides:

x = 6/7 + (2 i sqrt(19))/7 or x - 6/7 = -(2 i sqrt(19))/7

Hint: | Look at the second equation: Solve for x.

Add 6/7 to both sides:

 x = 6/7 + (2 i sqrt(19))/7 or x = 6/7 - (2 i sqrt(19))/7

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.