Find the zeros of 2x^2 - 16x + 27 using the quadratic formula. be sure to simplify the expression. can someone show me how this is done.

For a quadratic of the form , we have the quadratic formula 
,
where a is the coefficient (number before the variable) of the squared term, b is the coefficient of the linear term, and c is the constant term.

So, given , we can get that , and . We substitute these numbers into the quadratic formula above.











This is our final answer.

If you've never seen the quadratic formula, you can derive it by completing the square for the general form of a quadratic. Note that the  symbol (read: plus or minus) represents the two possible distinct solutions, except for zero under the radical, which gives only one solution.

It’s d for sure i just did also that’s why

Answer: f(x) = -2x^3 +3x^2 +11x -6see attachedan infinite number. since the magnitude of the leading coefficient is not specified, it may be any negative number. (we have chosen the smallest magnitude integer that makes all coefficients be integers.)step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.