Решите уравнение:1) 2x(5x-7)=2x^2-5; 3)(x-5)^2=3x^2-x+14

1)2x(5x-7)=2x²-5; 10x²-14x=2x²-5 10x²-2x²-14x+5=0 8x²-14x+5=0 D=(b)²-4*a*c   {формула} D=(-14)²-4*5*8=196-160=36 [latex] x_{1,2} = \frac{14 \frac{+}{-} 6}{16} [/latex] [latex] x_{1}= 1,25[/latex] [latex] x_{2}=0,5 [/latex] 3)(x-5)²=3x²-x+14 x²-10x+25=3x²-x+14 x²-3x²-10x+x+25-14=0 -2x²-9x+11=0 D=(-9)²-4*(-2)*11=81+88=169 x₁=[latex] \frac{9 +13 }{-4} [/latex] x₁=-5,5 x₂=[latex] \frac{9-13}{-4} [/latex] x₂=-1,5