How do you solve #7x ^ { 2} + 37x - 30= 0#?

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Answer 1 · 2018-05-17T12:08:33

#x = 5/7 or x = -6#

This is a quadratic equation in the form of #ax^2+bx+c=0#.

First, you multiply #a# and #c#, in this case, #7# and #-30#, and get #-210#. Then you get factors of #ac# #(-210)# which when added give #b# #(37)#.

The factors are #42# and #-5#. You substitute this factors in place of #b#

#(7x^2+42x)(-5x-30)=0#

#7x(x+6)-5(x+6)=0#

#7x-5=0 =>x=5/7#

#x+6=0 => x=-6#

Answer 2 · 2018-05-17T12:12:44

Solution: #x =-6 , x=5/7#

# 7 x^2 +37 x -30 =0 # or

# 7 x^2 +42 x -5 x -30 =0 # or

# 7 x( x+6) x -5 (x +6) =0 # or

#( x+6) (7 x-5) =0:.# either # x+6=0 :. x = -6. # or

# 7 x - 5 =0 :. x = 5/7 #

Solution: #x =-6 , x=5/7# [Ans]