Question 607617
There are other ways to solve a quadratic equation, but using the quadratic formula always works if you are careful in doing your calculations.
 
SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC FORMULA:
A quadratic equation is an equation of the form
{{{ax^2+bx+c=0}}} where the coefficients b, and c are any number and a is not zero.
Its solutions are given by the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
To solve a quadratic equation by using the quadratic formula, you need to determine the values for a, b, and c in the equation you are solving. You then plug those values into the formula, and calculate very carefully.
 
FOR {{{5x^2+16x-84=0}}}, a=5, b=16, and c=-84,
so {{{x=(-16 +- sqrt( 16^2-4*5*(-84)))/(2*5)=(-16 +- sqrt(256+1680))/10=(-16 +- sqrt(1936))/10=(-16 +- 44)/10}}}
The answers are
{{{x=(-16+44)/10=28/10=highlight(14/5)}}} and {{{x=(-16-44)/10=-60/10=highlight(-6)}}} 
 
FOR {{{4x^2-12x+9=0}}}, a=4, b=-12, and c=9,
so {{{x=(-(-12) +- sqrt((-12)^2-4*4*9))/(2*4)=(12 +- sqrt(144-144))/8=(-16 +- sqrt(0))/8=(12 +- 0)/8=12/8=3/2}}}
The one and only answer is
{{{highlight(x=3/2)}}}
 
THE OTHER WAYS TO SOLVE THOSE EQUATIONS:
Factoring, with some work, we get {{{5x^2+16x-84=(5x-14)(x+6)}}} ,
so the equation {{{5x^2+16x-84=0}}} turns into
{{{(5x-14)(x+6)=0}}} with solutions when {{{5x-14=0}}} --> {{{x=14/5}}}
and when {{{x+6=0}}} --> {{{x=-6}}}
 
The factoring is much easier for {{{4x^2-12x+9=0}}}, because it is obvious that {{{4x^2-12x+9=(2x-3)^2}}} . It's the square of a binomial, one of the special products you are taught about in polynomials.
The only solution for {{{(2x-3)^2=0}}} is the value for x that makes
{{{2x-3=0}}} --> {{{2x-3=0}}} --> {{{2x=3}}} --> {{{x=3/2}}}.
 
FACTORING {{{5x^2+16x-84}}} :
If it can be factored, it will be
{{{5x^2+16x-84=(5x+d)(x+f)=5x^2+(d+5f)x+df)}}} so we know that:
{{{df=-84}}} so d and f have different signs, and
two factors (one positive and one negative) whose product is {{{5*df=-420=5*(-84)=(d)*(5f)}}} will be such that {{{d+5f=16}}}
There are many pairs of factors to 420 (we can worry about the signs later):
420=(1)(420)=(2)(210)=(3)(140)=(4)(105)=(5)(84)=(6)(70)=(7)(60)=(10)(42)=(12)(35)=(14)(30)=(15)(28)=(20)(21),
but the only pair that adds up to 16 (we consider sign now) is 30 and -14 (30-14=16).
So, before collecting like term, the product could have been
{{{5x^2+30x-14x-84}}} and that is easier to factor by grouping, as follows:
{{{5x^2+30x-14x-84=5x(x+6)-14(x+6)=(5x-14)(x+6)}}}