How do you solve #x^ { 2} + 4x - 7= - 5#?

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Answer 1 · 2017-09-02T14:35:15

#x~~+0.4495#
#x~~-0.4495# both to 4 decimal places

Write as #x^2+4x-2=0#

Compare to the standardised form of #y=ax^2+bx+c#

Where in this case: #a=1; b=4 and c=-2#

Given that #x=(-b+-sqrt(b^2-4ac))/(2a)# we have:

#x=(-4+-sqrt(4^2-4(1)-2))/(2(1))#

#x=-2+-sqrt(24)/2#

#x=-2+-(cancel(2)sqrt(6))/cancel(2)#

#x=-2+-sqrt6 larr" Exact answer"#

#x~~0.4495#
#x~~-0.4495# both to 4 decimal places

Tony B