The general form of a quadratic equation is: $$ax^{2}+bx+c=0$$

x: unknown
a, b, c: constants with a != 0.

The general solution of a quadratic equation is: $$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$$

The plus-minus symbol "±" indicates that the quadratic equation has two solutions (one using '+', the other using '-'): $$x_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$$ These two solutions are called the roots of the quadratic equation.

Geometrically, these roots represent the x values at which any quadratic parabola crosses the x-axis. $$y = ax^{2} + bx + c \quad --> \quad x_{1},x_{2} \quad when \quad y = 0$$


Our goal is to transform our equation to the following form: $$x² + 2xb + b²$$ As: $$x² + 2xb + b² = (x + b)²$$ With the idea to be able to apply the square root on x².

We start with the general form: $$ax^{2}+bx+c=0$$ We multiply each side by 4a in order to transform the first terms to the square (2ax)² $$4a^{2}x^{2}+4abx+4ac=0$$ We rearrange $$4a^{2}x^{2}+4abx=-4ac$$

If we say X = 2ax (for reading purpose): $$4a^{2}x^{2}+4abx=X^{2} + 2Xb$$ b² is the only term missing to reach our form: x² + 2xb + b²,

So we add b² to both sides (It is also called completing the square) $$4a^{2}x^{2}+4abx+b^{2}=b^{2}-4ac$$ We use our formula to transform the left part $$(2ax+b)^{2}=b^{2}-4ac$$

Great, Now we may use the square root operator.

We take the square root of both sides $$2ax+b=\pm {\sqrt {b^{2}-4ac}}$$ We isolate x $$2ax=-b\pm {\sqrt {b^{2}-4ac}}$$ $$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$$

Proof done!


  • $$x*x=x²$$
  • $$x² = a \: (a > o) \quad ⇒ \quad x_1 = \sqrt{a}, \quad x_2 = -\sqrt{a}$$
  • $$(x + a)(x + b) = 0 \quad ⇒ \quad x_1 = -a, \quad x_2 = -b$$
  • $$(x + a)(x + b) = x² + (a + b)x + ab$$
  • $$(x + b)(x – b) = x² – b²$$
  • $$(x + b)² = x² + 2xb + b²$$