```Question 353963
First, let's solve {{{4y^2 - 12y - 19 = 0 }}}

Notice that the quadratic {{{4y^2-12y-19}}} is in the form of {{{Ay^2+By+C}}} where {{{A=4}}}, {{{B=-12}}}, and {{{C=-19}}}

Let's use the quadratic formula to solve for "y":

{{{y = (-(-12) +- sqrt( (-12)^2-4(4)(-19) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-12}}}, and {{{C=-19}}}

{{{y = (12 +- sqrt( (-12)^2-4(4)(-19) ))/(2(4))}}} Negate {{{-12}}} to get {{{12}}}.

{{{y = (12 +- sqrt( 144-4(4)(-19) ))/(2(4))}}} Square {{{-12}}} to get {{{144}}}.

{{{y = (12 +- sqrt( 144--304 ))/(2(4))}}} Multiply {{{4(4)(-19)}}} to get {{{-304}}}

{{{y = (12 +- sqrt( 144+304 ))/(2(4))}}} Rewrite {{{sqrt(144--304)}}} as {{{sqrt(144+304)}}}

{{{y = (12 +- sqrt( 448 ))/(2(4))}}} Add {{{144}}} to {{{304}}} to get {{{448}}}

{{{y = (12 +- sqrt( 448 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}.

{{{y = (12 +- 8*sqrt(7))/(8)}}} 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>)

{{{y = (12+8*sqrt(7))/(8)}}} or {{{y = (12-8*sqrt(7))/(8)}}} Break up the expression.

So the solutions in terms of y are {{{y = (12+8*sqrt(7))/(8)}}} or {{{y = (12-8*sqrt(7))/(8)}}}

But remember that {{{y=x^2}}}. So

{{{x^2 = (12+8*sqrt(7))/(8)}}} or {{{x^2 = (12-8*sqrt(7))/(8)}}}

Do you think you can take it from here?

If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=Algebra%20Help">jim_thompson5910@hotmail.com</a>

Also, feel free to check out my <a href="http://www.freewebs.com/jimthompson5910/home.html">tutoring website</a>

Jim
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