#1**0 **

To solve these types of equations, we usually have to isolate x. We do this by subtracting the right side from the left side.

\(100x^2 - x^2 + 20x - 8x + 1 - 16 = 0\)

\(99x^2 + 12x - 15 = 0\)

Why not divide by 3?

\(33x^2+4x-5=0\)

To factor the quadratic (assuming we can), we find two numbers that sum to 4 and multiply to (33 x -5). With some clever rearrangement, we get -11 and 15. Taking into account the leading coefficient and the factors of these two numbers, the quadratic factors as:

\((11x+5)(3x-1) = 0\)

Thus, the solutions occur when any of the terms are 0, which means \(x=-5/11, 1/3.\)

tinfoilhat Dec 9, 2021