Question

como resolver essa equação 2 ao quadrado -5 x -3 =0

Answer 1

$2x^{2} -5x-3=0= \Delta= b^{2} -4.a.c= \Delta= -5^{2} -4.2.(-3)= \Delta=25+24=49  x^{1} =-(-5)-+ \sqrt{49} /2.2=-5-7/4=-12/4=-3  x^{2} =-(-5)-+ \sqrt{49} /2.2=-5+7/4=2/4=$


S{-3,2/4}


Espero ter ajudado

Answer 2

$2x^{2}-5x-3=0 x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. x=\frac{- (-5)\pm\sqrt{(-5)^2-4*2*(-3)}}{2*2}. x=\frac{5\pm\sqrt{25-4*2*(-3)}}{4}. x=\frac{5\pm\sqrt{25+24}}{4}. x=\frac{5\pm\sqrt{49}}{4}. x=\frac{5\pm\{7}{4} x'= 3 x''= \frac{-2}{4} ou simplificando \frac{-1}{2}$
$S: [3, \frac{-1}{2}]$
Espero ter ajudado :)'