This week we will write a Java program that can solve the quadratic equation.

Recall the quadratic formula solves $a{x}^2+bx+c=0$ whenever $a0.$ In case no one has done you the proper service of deriving the formula, please allow me$.$

$a{x}^2+bx+c=0$

$Longrightarrow{x}^2+\frac{b}{a}x+\frac{c}{a}=0$

$Longrightarrow{x}^2+\frac{b}{a}x=-\frac{c}{a}$

$Longrightarrow{x}^2+\frac{b}{a}x+(\frac{b}{2a}{)}^2=-\frac{c}{a}+(\frac{b}{2a}{)}^2$

Longrightarrow$(x+\frac{b}{2a}{)}^2=-\frac{c}{a}+(\frac{b}{2a}{)}^2$

Longrightarrow$(x+\frac{b}{2a}{)}^2=\frac{b^2-4ac}{4a^2}$

Longrightarrow$+\frac{b}{2a}=+-\frac{\sqrt[\phantom{2}]{b-4ac}}{2a}$

Longrightarrowx$=-\frac{b}{2a}+-\frac{\sqrt[\phantom{2}]{b-4ac}}{2a}$

Longrightarrowx$=\frac{-b+-\sqrt[\phantom{2}]{b^2-4ac}}{2a}$

divide both sides of eq$.$ by a

subtracting constant term

completing the square

factoring left$-$hand side of equation

combine right$-$hand side terms

taking square root of both sides

subtracting constant term

et voil$!$

Notice that the term under the square root, $=b^2-4ac,$ says a lot about the solution. We call this term the discriminant.

If $>0,$ we are guaranteed two real solutions

Else if $=0,$ we have two solutions technically, but they are repeated

Else, it must be the case that $<mn\; id="EditorElement\_78223">0</mn>$ which implies there are two solutions that are complex numbers $($as they involve $i=\sqrt[\phantom{2}]{-1})$

I am providing a Java file darr that calculates the discriminant. You will be adding the logic that can store a user's input into the variables $a,b$ and $c.$

Given the sign of the discriminant, add the appropriate $"$if$",$ "else if$",$ "else" statements that tell the user what the solutions are. Here are a few examples: