solve the quadratic equation if an equaiton has not real roots states this, in ccase where the solutions incases where the solutions involve radicals, give both the radiccal form of the answer and a calculator approximation rounded two decimals. x^2=8x+9

Question

solve the quadratic equation  if an equaiton has not real roots states this, in ccase where the solutions incases where the solutions involve radicals, give both the radiccal form of the answer and a calculator approximation rounded two decimals.   x^2=8x+9

anonymous · Answer

$$x^2-8x=9$$
$$(x-4)^2=9+16=25$$
$$x-4=\pm5$$
$$x=4\pm 5$$
$$x=4+5=9$$ or 
$$x=4-5=-1$$

anonymous · Answer

Rewrite the quadratic equation  in the form :
$$ax ^{2}+bx+c=0$$  which implies
$$x ^{2}-8x-9=0$$
(x+1)(x-9)=0 (achieved by factoring the equation)
Therefore x=-1 and x=9
Moreover, the Discriminant, $$D=b ^{2}-4ac=100>0$$ indicating that the equation has two distinct real roots.