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Solutions (roots):

$$ x_{1,2} = \frac{b \pm \sqrt{b^2  4ac}}{2a} $$ 
If $ D = b^2  4ac $ is the discriminant , then the roots are
1. real and unique if $ D > 0 $
2. real and equal if $ D = 0 $
3. complex conjugate if $ D < 0 $
Let

$$ \begin{aligned} Q &= \frac{3a_2  a_1^2}{9} \\ R &= \frac{9a_1a_2  27a_3  2a_1^3}{54} \\ S &= \sqrt[\Large3]{R + \sqrt{Q^3 + R^2}} \\ T &= \sqrt[\Large3]{R  \sqrt{Q^3 + R^2}} \end{aligned} $$ 
Then solutions (roots) of the cubic equation are:

$$ \begin{aligned} x_1 &= S + T  \frac{1}{3}a_1 \\ x_2 &= \frac{1}{2} (S + T)  \frac{1}{3}a_1 + \frac{1}{2}\,i\,\sqrt{3}(ST) \\ x_3 &= \frac{1}{2} (S + T)  \frac{1}{3}a_1  \frac{1}{2}\,i\,\sqrt{3}(ST) \end{aligned} $$ 
If $D = Q^3 + R^2 $ is the discriminant of the cubic equation, then:
1. one root is real and two complex conjugate if $D > 0$
2. all roots are real and at last two are equal if $D=0$
3. all roots are real and unequal if $D < 0$
Let $y_1$ be a real root of the cubic equation

$$ y^3  a_2y^2 + (a_1a_34a_4)y+(4a_2a_4  a_3^2  a_1^2a_4) = 0 $$ 
Then solutions of the quartic equation are the 4 roots of

$$ z^2 + \frac{1}{2}\left(a_1 \pm \sqrt{a_1^2  4a_2+4y_1}\right)z + \frac{1}{2}\left(y_1 \pm \sqrt{y_1^2  4a_4}\right)= 0 $$ 
Please tell me how can I make this better.