 First use the discriminant to determine whether the equation has two nonreal com Falak Kinney 2021-10-02 Answered
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