How do you solve #6x ^ { 2} - 4x = 5#?

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Answer 1 · 2017-05-27T03:20:30

#x=(2+sqrt(34))/6,##(2-sqrt(34))/6#

Solve:

#6x^2-4x=5#

Subtract #5# from both sides.

#6x^2-4x-5=0#

This is a quadratic equation in standard form: #ax^2+bx+c#, where #a=6#, #b=-4#, and #c=-5#.

Use the quadratic formula to solve for #x#.

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

Insert the known values into the formula. Be mindful of the negatives.

#x=(-(-4)+-sqrt((-4)^2-4*6*(-5)))/(2*6)#

Simplify. Again, be mindful of the negatives.

#x=(4+-sqrt(16+120))/12#

#x=(4+-sqrt(136))/12#

Prime factorize #136#.

#x=(4+-sqrt(2xx2xx2xx17))/12#

Simplify the square root.

#x=(4+-sqrt(2^(2)xx2xx17))/12#

#x=(4+-2sqrt(34))/12#

Simplify the equation. Divide the numerator and denominator by #2#.

#x=(2+-sqrt(34))/6#

Solutions for #x#

#x=(2+sqrt(34))/6#

#x=(2-sqrt(34))/6#