How do you solve $5x^2+x-2=0$ using the quadratic formula?

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Answer 1 · 2016-07-30T22:13:49

${(-1-sqrt41)/10,(-1+sqrt 41)/10}$

$5x^2+x-2=0$

$y=ax^2+bx+c=0$

$"where ;"$

$a=5" ; "b=1" ; "c=-2$

$Delta=sqrt(b^2-4*a*c)$

$Delta=sqrt(1^2-4*5*(-2))$

$Delta=sqrt(1+4*5*2)$

$Delta=sqrt41$

$x_(1,2)=(-b±Delta)/(2a)$

$x_1=(-1-sqrt 41)/(2*5)$

$x_1=(-1-sqrt41)/10$

$x_2=(-1+sqrt 41)/(2*5)$

$x_2=(-1+sqrt41)/10$

How do you solve 5x^2+x-2=05x^2+x-2=0 using the quadratic formula?