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Find all real solutions to $x^2+ 4x+ 4 = 0$ using the quadratic formula.
Find the discriminant. Hint: The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is $\Delta = b^2 - 4ac$.
$\Delta = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{54px}}~} {}^2 - 4 \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A2}}\hspace{54px}}~} \times \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A3}}\hspace{54px}}~}$ $\Delta = 4 {}^2 - 4 \times 1 \times 4$
$\Delta = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B1}}\hspace{54px}}~} - \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{B2}}\hspace{54px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BR}}\hspace{54px}}~}$ $\Delta = 16 - 16 = 0$
Hint: If $\Delta > 0$ there are two real solutions, if $\Delta = 0$ there is only one, and if $\Delta < 0$ there are none.
So the number of real solutions is ${\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{noOfSolutions}}\hspace{54px}}~}}$. So the number of real solutions is ${1}$.
Substitute into the formula. Hint: The quadratic formula states $x = \dfrac{-b \pm \sqrt{\Delta}}{2a}$.
$x = \dfrac{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{54px}}~} \pm \sqrt{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C2}}\hspace{54px}}~}}}{2 \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C3}}\hspace{54px}}~}}$ $x = \dfrac{-4 \pm \sqrt{0}}{2 \times 1}$
Hint: $x = \dfrac{-b \pm 0}{2a}$.
$x = \dfrac{-4 \pm \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D1}}\hspace{54px}}~}}{\class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{D2}}\hspace{54px}}~}}$ $x = \dfrac{-4 \pm 0}{2}$
Therefore,
$x = \dfrac{\class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{E1}}\hspace{54px}}~}}{2} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{F1}}\hspace{54px}}~}$ $x = \dfrac{-4}{2} = -2$