Question

1-determine as raízes da equação do 2° grau
x²-5x-4???

Answer 1

$x^2-5x-4=0\\\\a=1\ \ \ \ b=-5\ \ \ \ c=-4\\\\\triangle=b^2-4ac\\\triangle=(-5)^2-4\cdot(1)\cdot(-4)\\\triangle=25+16\\\triangle=31\\\\\\x= \dfrac{-b \frac{+}{-} \sqrt{\triangle} }{2a}\\\\\\x'= \dfrac{5+ \sqrt{31} }{2}\\\\\\x"= \dfrac{5- \sqrt{31} }{2}$