How do you evaluate #sqrt(b^2-4ac)# for a=2, b=-5, c=2?

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Answer 1 · 2017-02-28T20:10:39

#+-3#

#sqrt(b^2-4ac)" "->" "sqrt((-5)^2-4(2)(2))#

#sqrt(+25-16)=sqrt(9)=+-3#

Answer 2 · 2017-02-28T20:14:35

See the entire solution process below:

To solve this you must substitute:

#color(red)(2)# for #color(red)(a)# in the expression.

#color(blue)(-5)# for #color(blue)(b)# in the expression.

#color(purple)(2)# for #color(purple)(c)# in the expression.

#sqrt(color(blue)(b)^2 - 4color(red)(a)color(purple)(c))# becomes:

#sqrt(color(blue)(-5)^2 - (4xxcolor(red)(2)xxcolor(purple)(2)))#

#sqrt(25 - 16)#

#sqrt(9) = +-3#