Is the solution shown below correct? Explain.

9x+2=8x2+6x

Negative 8 x squared + 3 x + 2 = 0. x = StartFraction negative 3 plus-or-minus StartRoot (3) squared minus (4) (negative 8) (2) EndRoot Over negative 16 EndFraction. x = StartFraction negative 3 plus-or-minus StartRoot 9 minus (64) EndRoot Over negative 16 EndFraction. x = StartFraction 3 plus-or-minus StartRoot 55 Endroot i Over 16 EndFraction.

Sample Response/Explanation: No. The correct values of a, b, and c were substituted in, but the formula was simplified wrong. The 64 should be added so the radicand is 73. There should be 2 real roots.

It is wrong

Step-by-step explanation:

What we have is as follows;

9x + 2 = 8x^2 + 6x

Bringing 9x + 2 yo the right hand side, we have;

8x^2 + 6x - 9x - 2 = 0

8x^2 - 3x - 2 = 0

So what is used here is the quadratic formula and we want to try to see if it was used correctly.

The form we have in the question is however;

-8x^2 + 3x + 2 = 0

which is obtained by taking the right hand side of the equation to the left

Now;

using the quadratic formula;

x = - b ± √ (b^2 - 4ac) / 2a

where a is - 8

b = 3

c = 2

Thus we have

-3 ± √ (9 - 4 (-8) 2) / 2 (-8)

-3 ± √ (9 + 64) / (-16)

So where is wrong is in the square root

what we are supposed to have is 9 + 64 = 73 and not otherwise