How do you solve $35x^ + 98x + 40=0$ ?

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Answer 1 · 2015-11-11T13:17:51

Assuming the intended equation was $35x^2+98x+40=0$ , use the quadratic formula to find:

$x=-7/5+-sqrt(1001)/35$

$35x^2+98x+40$ is of the form $ax^2+bx+c$ with $a=35$ , $b=98$ and $c=40$ .

This has zeros given by the quadratic formula:

$x=(-b+-sqrt(b^2-4ac))/(2a)=(-98+-sqrt(98^2-(4*35*40)))/(2*35)=-7/5+-sqrt(49^2-1400)/35$

$=-7/5+-sqrt(1001)/35$

How do you solve 35x^ + 98x + 40=035x^ + 98x + 40=0?