How do you solve #(5s + 7) ( s + 5) = 23#?

Question

Ethan Adkins · Accepted Answer

We have two imaginary solutions :

#s =(-7+sqrt(-691))/10=(-7+isqrt(691) )/10= -0.7000+2.6287i#

or:

#s =(-7-sqrt(-691))/10=(-7-isqrt(691) )/10= -0.7000-2.6287i# #(5s + 7) ( s + 5) = 23# Use BODMAS rule. First solve brackets using distributive property: #5s (s +5) + 7( s + 5) = 23# # 5s^2 + 25 +7s + 35 =23# # 5s^2 +7s + 60 =23# # 5s^2 +7s + 60 -23= 0# # 5s^2 +7s + 60 -23= 0# # 5s^2 +7s + 37= 0# According to the Quadratic Formula, s , the solution for #as^2+bs+c = 0# , where a, b and c are coefficients, is given by : #s = - b ± sqrt( b2-4ac)/(2a)# In our case, # a = 5# # b = 7# #c = 37 # Accordingly, #(b^2 - 4ac) = 49 - 740# #= -691# Applying the quadratic formula : #s = -7 ± sqrt( -691 )/10# #i # and # -i # are the square roots of minus 1 Accordingly, #sqrt( -691) = sqrt (691 imes(-1)) # # = sqrt (691 ) imes sqrt( -1) = ± sqrt (691) i # #sqrt(691)# when rounded to 4 decimal digits, =# 26.2869# Now, # s = ( -7 ± 26.287 i ) / 10# We have two imaginary solutions : #s =(-7+sqrt(-691))/10=(-7+isqrt(691) )/10= -0.7000+2.6287i# or: #s =(-7-sqrt(-691))/10=(-7-isqrt(691) )/10= -0.7000-2.6287i#

How do you solve #(5s + 7) ( s + 5) = 23#?

= sqrt (691 )\times sqrt( -1) =