How do you solve #4r ^ { 2} + 4r - 8= - 5#?

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Answer 1 · 2018-03-10T02:14:16

#r=-3/2,1/2#

Solve:

#4r^2+4r-8=-5#

Add #5# to both sides.

#4r^2+4r-8+5=0#

Simplify.

#4r^2+4r-3=0# is a quadratic equation in standard form:

#ax^2+bx+c=0,#

where:

#a=4,# #b=4,# #c=-3#

You can solve this equation by using the quadratic formula.

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

Substitute #r# for #x#, plug in the known values and solve.

#r=(-4+-sqrt(4^2-4*4*-3))/(2*4)#

#r=(-4+-sqrt(64))/8#

Simplify #sqrt64# to #8#.

#r=(-4+-8)/8#

#r=(-4+8)/8,# #(-4-8)/8#

#r=4/8,# #-12/8#

Reduce.

#r=1/2,# #-3/2#