# First use the discriminant to determine whether the equation has two nonreal com

First use the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real solutions. Then solve the equation.
$$\displaystyle{x}^{{{2}}}-{3}{x}-{54}={0}$$

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Step 1
Given : The quadratic equation $$\displaystyle{x}^{{{2}}}-{3}{x}-{54}={0}$$
we have to tell using the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real solutions also we have to solve the given equation.
Step 2
as we know that the discriminant is given as $$\displaystyle{b}^{{{2}}}-{4}{a}{c}$$
here $$\displaystyle{x}^{{{2}}}-{3}{x}-{54}={0}$$
a=1, b=-3, c=-54
$$\displaystyle{D}={\left(-{3}\right)}^{{{2}}}-{4}{\left({1}\right)}{\left(-{54}\right)}$$
$$\displaystyle{D}={\left(-{3}\right)}^{{{2}}}-{4}{\left({1}\right)}{\left(-{54}\right)}$$
D=9+216
D=225
since D is a perfect square thereforeThe roots exists are real and distinct.
$$\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{D}}}}}{{{2}{a}}}}$$
$$\displaystyle{x}={\frac{{-{\left(-{3}\right)}\pm\sqrt{{{225}}}}}{{{2}}}}$$
$$\displaystyle{x}={\frac{{{3}\pm{15}}}{{{2}}}}$$
$$\displaystyle{x}={\frac{{{3}+{15}}}{{{2}}}},{\frac{{{3}-{15}}}{{{2}}}}$$
$$\displaystyle{x}={\frac{{{18}}}{{{2}}}},-{\frac{{{12}}}{{{2}}}}$$
x=9, -6