How do you solve #b^ { 2} + 5b - 3= 0#?

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Answer 1 · 2017-05-26T17:41:12

#color(blue)(b=(-5+sqrt37)/2# or #color(blue)(b=(-5-sqrt37)/2#

#:.a=1,b=5,c=-3.#

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

#:.b=-(5)+-sqrt((5^2)-4(1)(-3))/(2(1))#

#:.b=(-5+-sqrt(25+12))/2#

#:.b=(-5+-sqrt37)/2#

#:.color(blue)(b=(-5+sqrt37)/2#

or

#:.color(blue)(b=(-5-sqrt37)/2#