Question 729502
You tried to use
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ,
the quadratic formula.
That cumbersome formula always works, but I only use it if I have to.
When you apply it properly, you get
{{{x = (-(-6) +- sqrt((-6)^2-4*4*2 ))/(2*4) }}} , which simplifies to
{{{x = (6 +- sqrt(36-32 ))/8 }}}
You had two errors:
{{{(-6)^2=(-6)(-6)=36}}} Products of two negative numbers are positive, so all squares are always positive
{{{4*4*2=16*2=32}}} I really do not know where your 72 came from.
So,
{{{x = (6 +- sqrt(36-32 ))/8 }}} --> {{{x = (6 +- sqrt(4))/8 }}} --> {{{x = (6 +- 2)/8 }}}
That gives you solutions
{{{x=(6-2)/8=4/8=1/2}}} and {{{x=(6+2)/8=8/8=1}}} .
 
NOTES:
Alternate ways to solve the problem are factoring, and completing the square.
You could have chosen to simplify the problem first and solve it by factoring, as ankor showed you better than what I was typing.
Factoring does not always work, and completing the square is not always easy.
However, the alternate ways do not require to memorize a formula, and sometimes they are much easier than applying the quadratic formula.
 
Factoring:
I had typed my wordy explanation, but ankor did it better.
 
Completing the square:
{{{(2x-3/2)^2=(2x)^2-2(2x)(3/2)+(3/2)^2=4x^2-3x+9/4}}} shares all the same terms in x as {{{4x^2-6x+2}}}
{{{4x^2-6x+2=0}}} --> {{{4x^2-6x=-2}}}
Adding {{{9/16}}} to both sides of the equal sign, we get a "completed" square on the left side:
{{{4x^2-6x+9/16=-2+9/4}}} --> {{{4x^2-6x+9/16=1/4}}} --> {{{(2x-3/2)^2=1/4}}}
Since {{{(1/2)^2=1/4}}} and {{{(-1/2)^2=1/4}}} it could be that
{{{2x-3/2=1/2}}} or that {{{2x-3/2=-1/2}}} and each of those equations leads us to a solution.