Question 201620
To use the quadratic formula we must first get the equation into the form:
{{{ax^2 + bx + c = 0}}}
This means:<ol><li>Get one side equal to zero</li><Change any subtractions on the non-zero side to additions of the opposites</li><li>Use the Commutative property of addition to rearrange the non-zero side in order from highest exponent to lowest.</li></ol>
Using these steps on your equation: {{{3n^2 - 15 = -4n}}}
1. Make one side zero. By adding 4n to both sides the right side will become zero:
{{{3n^2 - 15 + 4n = 0}}}
2. Change subtractions to additions
{{{3n^2 + (-15) + 4n = 0}}}
3. Rearrange
{{{3n^2 + 4n + (-15) = 0}}}<br>
Now we have the equation in the proper form for the quadratic formula:
{{{x = (-b +- sqrt(b^2 - 4ac))/2a}}}
Looking at your rearranged equation we should be able to tell that
a = 3
b = 4
c = -15
Substituting these into the appropriate places in the formula we get:
{{{x = ( -(4) +- sqrt((4)^2 - 4(3)(-15)))/2(3)}}}
Squaring the 4
{{{x = ( -4 +- sqrt(16 - 4(3)(-15)))/2(3)}}}
Multiplying
{{{x = ( -4 +- sqrt(16 - (-180)))/6}}}
Simplifying inside the square root
{{{x = ( -4 +- sqrt(196))/6}}}
{{{sqrt(196) = 14}}} Substituting we get
{{{x = (-4 +- 14)/6}}}
Splitting the +- we get
{{{x = (-4 + 14)/6}}} or {{{x = (-4 - 14)/6}}}
Simplifying each:
{{{x = 10/6 = 1&4/6 = 1&2/3}}} or {{{x = -18/6 = -3}}}
The solutions:
{{{x = 1&2/3}}} or {{{x = -3}}}