Question 996252
First the quadratic formula
{{{(-b +- sqrt(b^2-4ac))/2a}}}
Our equation is {{{x^2+12=0}}}
The a in the formula is the coefficient of the square term
so for our equation, a = 1
Since our equation does not have a middle term, b = 0
c is the value of the constant which for us is 12
Substituting we get
{{{(-0 +- sqrt(0-4(1)(12)))/2(1)}}}
{{{(-0 +- sqrt(-48))/2}}}
{{{(-0 +- sqrt(-12*4))/2}}}
{{{(-0 +- 2*sqrt(-12))/2}}}
{{{(-0 +- sqrt(-12))}}}
{{{(+- sqrt(-12))}}}
Since we have the square root of a negative number we have no real root
Notice that we could have solved more easily.
Take our equation, {{{x^2+12=0}}}
add -12 to each side
{{{x^2=-12}}}
Take the square root of each side and we get
{{{x=+-sqrt(-12)}}}