Question

(3x-5)^2-(3x-5)(4x+3)+9x^2-25=0

Answer 1

${(3x - 5)}^{2} - (3x - 5) \times (4x + 3) + {9x}^{2} - 25 = 0 \\ {9x}^{2} - 30x + 25 - ( {12x}^{2} + 9x- 20x - 15) + {9x}^{2} - 25 = 0 \\ {9x}^{2} - 30x + 25 - ( {12x}^{2} - 11x - 15)+ {9x}^{2} - 25 = 0 \\ {9x}^{2} - 30x + 25 - {12x}^{2} + 11x + 15+ {9x}^{2} - 25 = 0 \\ {6x}^{2} - 19x + 15 = 0 \\ \\ x = \frac{ - ( - 19) + - \sqrt{ {( - 19)}^{2} - 4 \times 6 \times 15} }{2 \times 6} \\ x = \frac{19 + - \sqrt{361 - 360} }{12} \\ x = \frac{19 + - \sqrt{1} }{12} \\ x = \frac{19 + - 1}{12} \\ \\ {x}^{1} = \frac{19 + 1}{12} = \frac{20}{12} = \frac{5}{3} \\ {x}^{2} = \frac{19 - 1}{12} = \frac{18}{12} = \frac{3}{2}$