How do you solve $35x^2+6x-9=0$ using the quadratic formula?

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Answer 1 · 2016-04-23T18:54:08

$x_1=-6/10$
$x_2=3/7$

$"if an equation is given as "ax^2+bx+c=0;$
$"the quadratic formula is determined b y "x_"1,2"=(-b±sqrt(b^2-4*a*c))/(2*a)$

$35x^2+6x-9=0$
$a=35" ;"b=6" ;"c=-9$

$Delta=sqrt(b^2-4*a*c)=sqrt(6^2+4*35*9)$
$Delta=sqrt(36+1260)=sqrt1296$
$Delta=36$

$x_1=(-b-Delta)/(2*a)" "x_1=(-6-36)/(2*35)" "x_1=-42/70$

$x_1=-6/10$

$x_2=(-b+Delta)/(2*a)" "x_2=(-6+36)/(2*35)" "x_2=30/70" "x_2=3/7$

How do you solve 35x^2+6x-9=0 35x^2+6x-9=0 using the quadratic formula?