```Question 183988
I'll do the first one to get you started

# 1

{{{2x^2-12x+14=0}}} Add 14 to both sides.

Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-12}}}, and {{{c=14}}}

Let's use the quadratic formula to solve for x

{{{x = (-(-12) +- sqrt( (-12)^2-4(2)(14) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-12}}}, and {{{c=14}}}

{{{x = (12 +- sqrt( (-12)^2-4(2)(14) ))/(2(2))}}} Negate {{{-12}}} to get {{{12}}}.

{{{x = (12 +- sqrt( 144-4(2)(14) ))/(2(2))}}} Square {{{-12}}} to get {{{144}}}.

{{{x = (12 +- sqrt( 144-112 ))/(2(2))}}} Multiply {{{4(2)(14)}}} to get {{{112}}}

{{{x = (12 +- sqrt( 32 ))/(2(2))}}} Subtract {{{112}}} from {{{144}}} to get {{{32}}}

{{{x = (12 +- sqrt( 32 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}.

{{{x = (12 +- 4*sqrt(2))/(4)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)

{{{x = (12)/(4) +- (4*sqrt(2))/(4)}}} Break up the fraction.

{{{x = 3 +- sqrt(2)}}} Reduce.

{{{x = 3+sqrt(2)}}} or {{{x = 3-sqrt(2)}}} Break up the expression.

So the solutions are {{{x = 3+sqrt(2)}}} or {{{x = 3-sqrt(2)}}}
```