Question 346644
How do I solve this quadratic equation?

  x^2 +4x - 6


It is not an equation yet, it is a trinomial (a polynomial with 3 terms).
To solve it for x set it to 0 (zero).


x^2 + 4x - 6 = 0
2 ways to solve this, either by factoring or by quadratic formula.


factoring and then testing with FOIL (First Outer Inner Last):
-6 = -1 * 6 = 1 * -6 = -2 * 3 = 2 * - 3 (4 possibilities of factors)
(x - 1)(x + 6) --> outer and inner add to 5, so not work
(x + 1)(x - 6) --> outer and inner add to -5, so not work
(x - 2)(x + 3) --> outer and inner add to 1, so not work
(x + 2)(x - 3) --> outer and inner add to -1, so not work


we are going to have to use the quadratic formula:
x^2 + 4x - 6 = 0 is of the form ax^2 + bx + c = 0,
where a = 1, b = 4, and c = -6
quadratic formula is:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
plug in values and solve:


{{{x = (-4 +- sqrt( 4^2 - 4*1*-6 ))/(2*1) }}}
{{{x = (-4 +- sqrt( 16 - 4*-6 ))/2 }}}
{{{x = (-4 +- sqrt( 16 - (-24) ))/2 }}}
{{{x = (-4 +- sqrt( 16 + 24 ))/2 }}}
{{{x = (-4 +- sqrt( 40 ))/2 }}}
{{{x = (-4 +- sqrt( 4*10 ))/2 }}}
{{{x = (-4 +- 2*sqrt( 10 ))/2 }}}
{{{x = -2 +- sqrt( 10 ) }}}
{{{x1 = -2 + sqrt( 10 ) }}}
{{{x2 = -2 - sqrt( 10 ) }}}


x = -2 + sqrt(10) = 1.1623 rounded to 4 places
or x = -2 - sqrt(10) = -5.1623 rounded to 4 places


check:
x^2 + 4x - 6 = 0
(-2 + sqrt(10))^2 + 4(-2 + sqrt(10)) - 6 = 0
4 + 2 * -2 * sqrt(10) + 10 - 8 + 4sqrt(10) - 6 = 0
4 - 4sqrt(10) + 2 + 4srt(10) - 6 = 0
6 - 6 = 0
0 = 0 , yes
(-2 - sqrt(10))^2 + 4(-2 - sqrt(10)) - 6 = 0
4 + 2 * -2 * -sqrt(10) + 10 - 8 - 4sqrt(10) - 6 = 0
4 + 4sqrt(10) + 2 - 4sqrt(10) - 6 = 0
6 - 6 = 0
0 = 0, yes