Answer:
A quadratic equation \(ax^2+bx+c = 0\) ....[1] where, a, b and c are coefficient.
Then the solution is given by:
\(x = -\frac{b \pm \sqrt{b^2-4ac}}{2a}\) ....[2]
As per the statement:
Given the equation:
\(x^2=5-x\)
⇒\(x^2+x-5 = 0\)
On comparing with [1] we have;
a =1, b = 1 and c = -5
Then substitute these in [2] we have;
\(x = \frac{-1 \pm \sqrt{1^2-4(1)(-5)}}{2(1)} =\frac{-1 \pm \sqrt{1+20}}{2} =\frac{-1 \pm \sqrt{21}}{2}\)
Therefore, the values of x are:
\(x = \frac{-1 + \sqrt{21}}{2}\) , \(\frac{-1 - \sqrt{21}}{2}\)
