First we render the quartic into monic form: :$x^4 + \dfrac b a x^3 + \dfrac c a x^2 + \dfrac d a x + \dfrac e a = 0$ Completing the square in $x^2$: :$\paren {x^2 + \dfrac b {2 a} x}^2 + \paren {\dfrac c a - \dfrac {b^2} {4 a^2} } x^2 + \dfrac d a x + \dfrac e a = 0$ Then we introduce a new variable $y$: :$\paren {x^2 + \dfrac b {2 a} x + \dfrac y 2}^2 + \paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y} x^2 + \paren {\dfrac d a - \dfrac b {2 a} y} x + \paren {\dfrac e a - \dfrac {y^2} 4} = 0$ This equation is valid for any $y$, so let us pick a value of $y$ so as to make: :$\paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y} x^2 + \paren {\dfrac d a - \dfrac b {2 a} y} x + \paren {\dfrac e a - \dfrac {y^2} 4}$ have a zero discriminant. That is: :$\paren {\dfrac d a - \dfrac b {2 a} y}^2 = 4 \paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y} \paren {\dfrac e a - \dfrac {y^2} 4}$ After some algebra, this can be expressed as a cubic in $y$: :$y^3 - \dfrac c a y^2 + \paren {\dfrac {b d} {a^2} - \dfrac {4 e} a} y + \paren {\dfrac {4 c e} {a^2} - \dfrac {b^2 e} {a^3} - \dfrac {d^2} {a^2} } = 0$ Using (for example) Cardano's Formula, we can find a real solution of this: call it $y_1$. Now a quadratic equation $p x^2 + q x + r$ can be expressed as: :$p \paren {\paren {x + \dfrac q {2 p} }^2 - \dfrac {q^2 - 4 p r} {4 p^2} }$ If that quadratic has a zero discriminant, i.e. $q^2 = 4 p r$, then this reduces to: :$p \paren {\paren {x + \dfrac q {2 p} }^2}$ which in turn becomes: :$p \paren {\paren {x + \pm \sqrt {\dfrac r p} }^2}$ as $q^2 = 4 p r \implies \dfrac {q^2} {4 p^2} = \dfrac r p$. So, as: :$\paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y_1} x^2 + \paren {\dfrac d a - \dfrac b {2 a} y_1} x + \paren {\dfrac e a - \dfrac {y_1^2} 4}$ has a zero discriminant (we picked $y_1$ to make that happen), we can write it as: :$\paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y_1} \paren {x \pm \dfrac {\sqrt {\paren {\dfrac e a - \dfrac {y_1^2} 4} } } {\sqrt {\paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y_1} } } }^2$ Now we return to the equation: :$\paren {x^2 + \dfrac b {2 a} x + \dfrac {y_1} 2}^2 + \paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y_1} x^2 + \paren {\dfrac d a - \dfrac b {2 a} y_1} x + \paren {\dfrac e a - \dfrac {y_1^2} 4} = 0$ which can now be written: :$\paren {x^2 + \dfrac b {2 a} x + \dfrac {y_1} 2}^2 = \paren {\dfrac {b^2} {4 a^2} - \dfrac c a + y_1} \paren {x \mp \dfrac {\sqrt {\paren {\dfrac {y_1^2} 4 - \dfrac e a} } } {\sqrt {\paren {\dfrac {b^2} {4 a^2} - \dfrac c a + y_1} } } }^2$ Taking square roots of both sides: :$x^2 + \dfrac b {2 a} x + \dfrac {y_1} 2 = \pm x \sqrt {\paren {\dfrac {b^2} {4 a^2} - \dfrac c a + y_1} } \mp \sqrt {\dfrac {y_1^2} 4 - \dfrac e a}$ Arranging into canonical quadratic form: :$(1): \quad x^2 + \paren {\dfrac b {2 a} \pm \dfrac 1 2 \sqrt {\dfrac {b^2} {a^2} - \dfrac {4 c} a + 4 y_1} } x + \dfrac 1 2 \paren {y_1 \mp \sqrt {y_1^2 - \dfrac {4 e} a} } = 0$ Let: :$p = \dfrac b a \pm \sqrt {\dfrac {b^2} {a^2} - \dfrac {4 c} a + 4 y_1}$ :$q = y_1 \mp \sqrt {y_1^2 - \dfrac {4 e} a}$ Then equation $(1)$ can be written as: :$x^2 + \dfrac p 2 x + \dfrac q 2 = 0$ Using the Quadratic Formula, putting $a = 1, b = \dfrac p 2, c = \dfrac q 2$: {{begin-eqn}} {{eqn | l = x       | r = \dfrac {-\dfrac p 2 \pm \sqrt {\dfrac {p^2} 4 - 4 \dfrac q 2} } 2       | c =  }} {{eqn | r = \dfrac {-\dfrac p 2 \pm \sqrt {\dfrac 1 4} \sqrt {p^2 - 8 q} } 2       | c =  }} {{eqn | r = \dfrac {-p \pm \sqrt {p^2 - 8 q} } 4       | c =  }} {{end-eqn}} Hence the result. {{qed}}