What are the solutions to #x^2 = 14x - 40#?

1 Answer
Bruno Pedron
Nov 13, 2015

#x'=10#
#x''=4#

Explanation:

In order to use Bhaskara's formula, the expression must be equal to zero. Therefore, change the equation to:
#x^2-14x+40=0#,
The apply the formula:

#(-b+-sqrt(b^2-4ac))/(2a)#, where a is the number that multiplies the quadratic term, b is the number that multiplies #x# and c is the independent term.

#(14+-sqrt(14^2-4*(1*40)))/(2*1)=(14+-sqrt(36))/2=(14+-6)/2=7+-3#

Solving for x':
#x'=7+3=10#

Solving for x'':
#x''=7-3=4#,

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