Rozwiązanie :
[tex](2-3x)^2=4x(4x-2)\\\\2^2-2\cdot2\cdot3x+(3x)^2=16x^2-8x\\\\9x^2-12x+4=16x^2-8x\\\\9x^2-12x+4-16x^2+8x=0\\\\-7x^2-4x+4=0 \ \ |\cdot(-1)\\\\7x^2+4x-4=0\\\\a=7, \ b=4, \ c=-4\\\\\Delta=4^2-4\cdot7\cdot(-4)=16+112=128\\\\\sqrt{\Delta}=\sqrt{128}=\sqrt{64\cdot2}=8\sqrt2\\\\x_1=\frac{-4-8\sqrt2}{2\cdot7}=\frac{-4-8\sqrt2}{14}=\frac{-2-4\sqrt2}{7}\\\\x_2=\frac{-4+8\sqrt2}{2\cdot7}=\frac{-4+8\sqrt2}{14}=\frac{-2+4\sqrt2}{7}[/tex]
Wykorzystane wzory :
[tex](a-b)^2=a^2-2ab+b^2\\\\y=ax^2+bx+c\\\\\Delta=b^2-4ac\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}[/tex]
[tex](2-3x)^{2} = 4x(4x-2)\\\\4 - 12x + 9x^{2} = 16x^{2}-8x\\\\9x^{2}-16x^{2}-12x+8x+4 = 0\\\\-7x^{2}-4x+4 = 0 \ \ /\cdot(-1)\\\\7x^{2}+4x-4 = 0\\\\a = 7, \ b = 4, \ c = -4\\\\\Delta = b^{2}-4ac = 4^{2}-4\cdot7\cdot(-4) = 16 + 112 = 128\\\\\sqrt{\Delta } = \sqrt{128} = \sqrt{64\cdot2} = 8\sqrt{2}\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-4-8\sqrt{2}}{2\cdot7} = \frac{-4-8\sqrt{2}}{14} = \frac{-2-4\sqrt{2}}{7}\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-4+8\sqrt{2}}{14} = \frac{-2+4\sqrt{2}}{7}[/tex]