Question 116773
Given:
.
{{{3x^2+4x-15=0}}}
.
This is in the standard quadratic form of:
.
{{{ax^2 + bx + c = 0}}}
.
By comparing the standard form to the given problem, you can see that a = 3, b = 4, and c = -15
.
For the standard form, the values of x that satisfy the equation are given by:
.
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
.
So all you have to do to solve the given equation is to substitute 3 for a, 4 for b, and -15
for c into the equation for x. When you do, that equation becomes:
.
{{{x = (-(4) +- sqrt( 4^2-4*3*(-15) ))/(2*3) }}}
.
Multiply out the denominator 2*3 = 6 to make the equation become:
.
{{{x = (-(4) +- sqrt( 4^2-4*3*(-15) ))/6 }}}
.
Work inside the radical. {{{4^2 = 16}}} and {{{-4*3*(-15) = 180}}} and substituting 
these values results in:
.
{{{x = (-(4) +- sqrt( 16+180))/6 }}}
.
Combine the terms in the radical:
.
{{{x = (-(4) +- sqrt( 196))/6 }}}
.
But the square root of 196 is 14. Substituting this results in:
.
{{{x = (-(4) +- 14)/6 }}}
.
Note that -(4) = -4 which simplifies the equation to:
.
{{{x = (-4 +- 14)/6 }}}
.
So there are two possible values of x as follows:
.
{{{x = (-4 + 14)/6  = 10/6 = 5/3}}}
.
and
.
{{{x = (-4 - 14)/6 = -18/6 = -3}}}
.
Those are the two answers. Hope this helps to familiarize you with the quadratic equation and
how it can be used to solve quadratic equations. 
.