How do you solve #18x ^ { 2} - 60x + 50=0#?

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Answer 1 · 2018-01-25T02:20:44

#x=5/3#

Solve:

#18x^2-60x+50=0#

Factor the common #2# from the left-hand side.

#2(9x^2-30x+25)=0#

Divide both sides by #2#.

#(color(red)cancel(color(black)(2))^1(9x^2-30x+25))/color(red)cancel(color(black)(2))^1=0/2#

Simplify.

#9x^2-30x+25=0#

This is a quadratic equation in standard form:

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

where:

#a=9#, #b=-30#, and #c=25#.

Solve for #x# using the quadratic formula.

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

Plug in the known values.

#x=(-(-30)+-sqrt((-30)^2-4*9*25))/(2*9)#

Simplify.

#x=(30+-sqrt(900-900))/18#

#x=(30+-sqrt0)/18#

#x=(30+-0)/18#

#x=30/18#

Reduce by dividing the numerator and denominator by #6#.

#x=(30-:6)/(18-:6)#

#x=5/3#