Question

preciso urgente da resposta desse calculo usando o delta (x-1) (2x+3)=0

Answer 1

$(x-1) (2x+3)=0\\\\\\ 2 x^{2} +3x-2x-3=0\to \\\\ 2 x^{2} +x-3=0\\\\ a=2;b=1;c=-3\\\\\\ \Delta=b^2-4ac\to \Delta=1^2-4*2*(-3)\to \Delta=1+24\to \Delta=25\\\\ x' \neq x''\\\\ x= \frac{-b\pm \sqrt{\Delta} }{2a} \to x= \frac{-1\pm \sqrt{25} }{2*2} \to x= \frac{-1\pm 5 }{4} \to\\\\ \\ x'= \frac{-1+ 5 }{4} \to x'= \frac{4 }{4} \to x'=1\\\\\\ x''= \frac{-1-5 }{4} \to x''= \frac{-6 }{4} \to x''=- \frac{3 }{2} \\\\\\ S=(- \frac{3 }{2};1)$


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