Remember solving quadractic equations? If you have a quadractic equation in "standard form," which is
$Ax^2+Bx+C=0$, then the two solutions to this equation will be
$$x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}.$$

In grade-school algebra the equations are kept simple in that the roots (or answers) to quadractic equations are always real. If you look carefully at the quadractic formula, there is a $\sqrt{B^2-4AC}$. As you know, you cannot take the square root of a negative number (if you want roots that are real numbers). In other words, $B^2-4AC$ (which is called the discriminant) must always be $\ge 0$, so in a quadratic equation solver, we must check that this is true (with an

In grade-school algebra the equations are kept simple in that the roots (or answers) to quadractic equations are always real. If you look carefully at the quadractic formula, there is a $\sqrt{B^2-4AC}$. As you know, you cannot take the square root of a negative number (if you want roots that are real numbers). In other words, $B^2-4AC$ (which is called the discriminant) must always be $\ge 0$, so in a quadratic equation solver, we must check that this is true (with an

`if`

statement) before continuing.
`if`

statement. You also have to complete
the `x1=`

and `x2=`

lines to give the $+$ and $-$ roots, as computed with the
quadratic formula above.
Type your code here:

See your results here:

This code will not run! What will you put in the

`if ???? then`

statement to
ensure that $d$ (the discriminant) is $\ge 0$ before continuing to compute
$x1$ and $x2$, the two solutions to the equation?
By the way, in this case we'll say that $x1=\frac{-B+\sqrt{B^2-4AC}}{2A}$ and $x2=\frac{-B-\sqrt{B^2-4AC}}{2A}.$