Question

me ajudem x(2x-1)+6=4(x+1)

(x-1)(x-2)=6

Answer 1

$x(2x-1)+6=4(x+1)\to \\\\ 2 x^{2} -x+6=4x+4\to\\\\ 2 x^{2} -x+6-4x-4=0\to \\\\2 x^{2} -5x+2=0\\\\ a=2;b=-5;c=2\\\\\\ \Delta=b^2-4ac\to \Delta=(-5)^2-4*2*2\to \Delta=25-16\to \Delta=9\\\\ x' \neq x''\\\\\\ x= \frac{-b\pm \sqrt{\Delta} }{2a} \to x= \frac{-(-5)\pm \sqrt{9} }{2*2} \to x= \frac{5\pm 3 }{4} \to \\\\\\x'= \frac{5+3 }{4} \to x'= \frac{8}{4} \to x'=2\\\\\\ x''= \frac{5-3 }{4} \to x''= \frac{2 }{4} \to x= \frac{1 }{2} \\\\\\\\ S=(\frac{1 }{2};2)$






$(x-1)(x-2)=6\to \\\\x^{2} -2x-x+2-6=0\to \\\\x^{2} -3x-4=0\\\\ a=1;b=-3;c=-4\\\\\\ \Delta=b^2-4ac\to \Delta=(-3)^2-4*1*(-4)\to \Delta=9+16\to \Delta=25\\\\ x' \neq x''\\\\\\x= \frac{-b\pm \sqrt{\Delta} }{2a} \to x= \frac{-(-3)\pm \sqrt{25} }{2*1} \to x= \frac{3\pm 5 }{2} \to\\\\\\ x'= \frac{3+ 5 }{2} \to x'= \frac{8 }{2} \to x'=4\\\\\\ x''= \frac{3-5 }{2} \to x''= \frac{-2 }{2} \to x''=-1\\\\\\\\ S=(-1;4)$