How do you solve #(x + 5) ( x - 3) = - 2( 2x + 7)#?

2 Answers
Meave60
May 7, 2017

#x=color(red)(-3+2sqrt10),color(blue)(-3-2sqrt10#

Explanation:

Solve:

#(x+5)(x-3)=-2(2x+7)#

Expand.

#(x+5)(x-3)=-4x-14#

FOIL.

#x^2+2x-15=-4x-14#

Add #4x# to both sides.

#x^2+6x-15=-14#

Add #14# to both sides.

#x^2+6x-1#

#x^2+6x-1# is a quadratic equation in standard form: #ax^2+bx+c#, where #a=1#, #b=6#, #c=-1#.

Use the quadratic formula to solve.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Substitute the given values into the formula.

#x=(-6+-sqrt(6^2-4xx1xx-1))/(2xx1)#

Simplify.

#x=(-6+-sqrt(36+4))/2#

#x=(-6+-sqrt(40))/2#

Prime decompose #sqrt(40)#.

#x=(-6+-sqrt(2xx2xx2xx5))/2#

#x=(-6+-sqrt(2^2xx2xx5))/2#

#x=(-6+-2sqrt10)/2#

Simplify.

#x=-3+-2sqrt10#

Solutions for #x#.

#x=color(red)(-3+2sqrt10),color(blue)(-3-2sqrt10#

Mr. Whoever
May 7, 2017

#x~~6.48#
#x~~ -4.48#

Explanation:

#(x+5)(x-3)= -2(2x+7)#
#x^2-3x+5x-15=4x+14#
#x^# (2+√120 ) / 2# (2+√120 ) / 2# (2+√120 ) / 2# (2+√120 ) / 2# =4x+14+3x-5x+15# #x^2=4x+3x-5x+14+15# #x^2=2x+29# #x^2-2x-29=0#

Apply this formula:
# (-b ±√b^2- 4ac) / (2a) #

#a=1#
#b=-2#
#c=-29#

# [- (-2) ±√(-2)^2 - (4×1× -29) ] / (2×1)#

# [2±√4 - (-116) ] / 2 #

# [2±√4+116 ] / 2#

# [ 2±√120 ] / 2 #

# (2+√120 ) / 2# or #(2-√120)/2#

#~~ (2+10.95) /2# #~~(2-10.95)/2#

#12.95/2 = 6.48# # -8.95/2 = -4.48#

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