How do you solve #r^ { 2} - 10r - 19= 0#?

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How do you solve #r^ { 2} - 10r - 19= 0#?

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Answer 1 · 2017-04-08T01:45:56

#r=(10\pmsqrt176)/2#

Explanation:

#r=(-b\pmsqrtb^2-4ac)/(2a)#
replace variables with numbers with a=1, b=-10, and c=-19
#r=-(-10)\pmsqrt((-10)^2-4(1)(-19))/(2(1))#
#r=(10\pmsqrt(100+76))/(2)#
#r=(10\pmsqrt176)/(2)#

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How do you solve #r^ { 2} - 10r - 19= 0#?

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