How to solve 2x-1/3 = x^2+2x/5 using the quadratic formula?

1 Answer
Jan 17, 2016

#x=(12+-sqrt69)/15#

Explanation:

I am interpreting the given equation as:

#2x-1/3=x^2+(2x)/5#

Move these all to the same side of the equation.

#0=x^2+(2x)/5-2x+1/3#

Simplify the terms with #x#.

#0=x^2+(2x)/5-(10x)/5+1/3#

#0=x^2-(8x)/5+1/3#

We could apply the quadratic formula to this, but it would be simpler if we eliminated the fractions. To do this, multiply both sides by #15#.

#0=15x^2-24x+5#

We can now use the quadratic formula, which states that for a quadratic function #0=ax^2+bx+c#, the roots are

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

Thus, #a=15,b=-24,c=5#, so

#x=(-(-24)+-sqrt((-24)^2-(4xx15xx5)))/(2xx15)#

#x=(24+-sqrt(576-300))/30#

#x=(24+-sqrt276)/30#

#x=(24+-sqrt(2^2xx69))/30#

#x=(24+-2sqrt69)/30#

#x=(12+-sqrt69)/15#