How do you solve #-7x ^ { 2} + 2x + 4= 0#?

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Answer 1 · 2018-01-20T14:45:01

#x~~-0.63,0.91#

The equation is in the form of #ax^2+bx+c=0# for #a!=0#, so we can use the quadratic formula, which states that

#x=(-b+-sqrt(b^2-4ac))/(2a)#

where there are two solutions for #x#.

Plugging in #a=-7#, #b=2#, and #c=4#, we get

#x=(-2+-sqrt(2^2-(4*-7*4)))/-14#

#x=(-2+-sqrt(4+112))/-14#

#xapprox(-2+-10.77)/-14#

#xapprox 8.77/-14,(-12.77)/-14#

#xapprox -0.63,0.91#