Question 164077
Three ways to solve a quadratic equation:
1) Use the "quadratic formula" and this will always work:{{{x = (-b+-sqrt(b^2-4ac))/2a}}}
2) "Completing the square", this will always work too.
3) "Factoring", this will work only if the equation is factorable.
Which one to use? Your experience in solving quadratic equations is probably the best guide, but I always try factoring first and, if I'm unable to factor, then I'll use the formula, but sometimes, depending on what the problem is asking for, I'll use completing the square.
Let's try these methods on your equation:
1) The quadratic formula.
To use this, your equation should be written in the "standard format":
{{{ax^2+bx+c = 0}}} and, in this case, it already is.
{{{12^x^2-10x-42 = 0}}} Here, a = 12, b = -10, and c = -42. Apply the formula:
{{{x = (-(-10)+-sqrt((-10)^2-4(12)(-42)))/2(12)}}} Simplify this.
{{{x = (10+-sqrt(100-(-2016)))/24}}}
{{{x = (10+-sqrt(2116))/24}}}
{{{x = (10+46)/24}}} or {{{x = (10-46)/24}}}
{{{x = 2.33}}} or {{{x = -1.5}}}
2) Completing the square.
{{{12x^2-10x-42 = 0}}} First, divide through by 12 to get the {{{x^2}}} coefficient equal to 1.
{{{x^2-(5/6)x-(7/2) = 0}}} Now add {{{7/2}}} to both sides.
{{{x^2-(5/6)x = 7/2}}} Complete the square in x by adding the square of half the x-coefficient, that's ({{{((1/2)(-5/6))^2 = 25/144}}}) to both sides.
{{{x^2-(5/6)x+25/144 = (25/144)+(7/2)}}} Rewrite the left side as the square of the binomial:{{{(x-5/12)}}} and simplify the right side.
{{{(x-(5/12))^2 = (25+504)/144}}}
{{{(x-(5/12))^2 = 529/144}}} Now take the square root of both sides.
{{{x-(5/12) = 23/12}}} or {{{x-(5/12) = -(23/12)}}} Add {{{5/12}}} to both sides.
{{{x = (5+23)/12}}} or {{{x-5 = (5-23)/12}}}
{{{x = 28/12}}} or {{{x = -18/12}}}
{{{x = 2.33}}} or {{{x = -1.5}}}
3) Factoring.
{{{12x^2-10x-42 = 0}}} Factor the trinomial:
{{{(4x+6)(3x-7) = 0}}} Apply the zero product rule:
{{{4x+6 = 0}}} or {{{3x-7 = 0}}} 
If {{{4x+6 = 0}}} then {{{4x = -6}}} so that {{{x = -6/4}}} or {{{x = -1.5}}}
If {{{3x-7 = 0}}} then {{{3x = 7}}} so that {{{x = 7/3}}} or {{{x = 2.33}}}