+133 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 \(x=-5/11, 1/3.\)
