(-x)cdot (x-43)=0 (3-x)cdot (x+4)=(1)

The first equation is already solved in the example. The solution is x = 0, x = 43.The second equation, , is a quadratic equation and requires a different approach than the simple zero-product property used in the examples. We need to expand and rearrange it into standard form:(3-x)(x+4) = 13x + 12 - x² - 4x = 1-x² - x + 12 = 1-x² - x + 11 = 0x² + x - 11 = 0This quadratic equation doesn't factor easily. We would need to use the quadratic formula to solve for x:x = [-b ± √(b² - 4ac)] / 2awhere a = 1, b = 1, and c = -11.Therefore:x = [-1 ± √(1² - 4 * 1 * -11)] / (2 * 1)x = [-1 ± √(1 + 44)] / 2x = [-1 ± √45] / 2x = [-1 ± 3√5] / 2So the two solutions are:x = (-1 + 3√5) / 2 and x = (-1 - 3√5) / 2In summary:* ** **: x = 0, x = 43* ** **: x = (-1 + 3√5) / 2, x = (-1 - 3√5) / 2