The line L with slope 2 intersects with hyperbola (x ^ 2) / 3 - (y ^ 2) / 2 = 1 at two points a and B, and the absolute value of AB is 4. The equation of line L is obtained y=2x+b. x²/3-（2x+b)²/2=1. 10x²+12bx+3b²+6＝0. |x1-x2|＝√（24b²-240）／10. |y1-y2|＝2√（24b²-240）／10. （x1-x2）²＋（y1-y2）²＝16. b²＝55／3.b＝±√165／3. The linear equation is: L1: y = 2x + √ 165 / 3 L2:y＝2x-√165／3. |x1-x2|＝√（24b²-240）／10. |y1-y2|＝2√（24b²-240）／10. This step is not clear! √（24b²-240）／10. How did it come about

The line L with slope 2 intersects with hyperbola (x ^ 2) / 3 - (y ^ 2) / 2 = 1 at two points a and B, and the absolute value of AB is 4. The equation of line L is obtained y=2x+b. x²/3-（2x+b)²/2=1. 10x²+12bx+3b²+6＝0. |x1-x2|＝√（24b²-240）／10. |y1-y2|＝2√（24b²-240）／10. （x1-x2）²＋（y1-y2）²＝16. b²＝55／3.b＝±√165／3. The linear equation is: L1: y = 2x + √ 165 / 3 L2:y＝2x-√165／3. |x1-x2|＝√（24b²-240）／10. |y1-y2|＝2√（24b²-240）／10. This step is not clear! √（24b²-240）／10. How did it come about

|AB|=SQR[（X1-X2)^2+（y1+y2）^2]
=√ {(x1-x2) ^ 2 * [1 + slope]}
=|X1-x2 | * √ (1 + slope)
The simultaneous equations eliminate the unknowns and get ax ^ 2 + BX ^ 2 + C = 0,
Find x1, X2, and then | x1-x2 | = (b ^ 2-4ac) / | a|
Remember this conclusion, it can help you a lot!