How do you solve #(2x ^ { 2} + 3x - 5) + ( x ^ { 2} + 11x - 1) = 180#?

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Answer 1 · 2018-07-18T08:45:41

#x_(1,2)=-7/3pmsqrt(607)/3#

Combining like Terms

#3x^2+14x-6=180#

adding #-180#

#3x^2+14x-186=0#

dividing by #3#

#x^2+14/3x-62=0#

so we get

by the quadratic formula
#x^2+px+q=0#

#x_(1,2)=-p/2pmsqrt(p^2/4-q)#

with #p=14/3,q=-162#

#x_(1,2)=-7/3pm\sqrt(607)/3#

Answer 2 · 2018-07-18T08:59:47

Collect like terms

#2x^2+3x-5+x^2+11x-1=180#

#3x^2+14x-6-180=0#

#3x^2+14x-186=0#

Use the formula

#x=(-14\pmsqrt(14^2+4xx3xx186))/(2xx3)#

#x=[-14\pmsqrt(2428)]/6#

#x=[-7+sqrt607]/3# or #x=[-7-sqrt607]/3#

#x=5.87912333 or x=-10.54579#