How do you solve #q^2-21q=-20#?

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Answer 1 · 2016-04-02T17:33:57

The solutions are:

# color(green)(q = 20#

# color(green)(q = 1 #

#q^2 - 21 q = -20#

#q^2 - 21 q + 20 = 0 #

The equation is of the form #color(blue)(aq^2+bq+c=0# where:

#a=1, b=-21, c=20#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (-21)^2-(4*1 * 20)#

# = 441 - 80 =361#

The solutions are normally found using the formula:

#q=(-b+-sqrtDelta)/(2*a)#

#q = (-( - 21)+-sqrt(361))/(2*1) = ((21+- 19))/2#

# q= ((21+ 19))/2 = 40 / 2 = 20#

# q= ((21 - 19))/2 = 2 / 2 = 1 #