If the highest power of a variable is 2, then we talk about quadratic equations. The solution is found by substituting multipliers into the formula.[br][br]First, the equation must be rewritten in the standard form[br][br]  [math]\large \textcolor{blue}{ax^2+bx+c=0.}[/math][br] [br]Solutions are[br][br] [math]\large \textcolor{blue}{x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.}[/math][br][br]If the discriminant (expression inside the square root) [math] b^2-4ac[/math] [br][br][list][*]> 0, there are two different real solutions. [/*][*] = 0, there is only one real solution. [/*][*] < 0, there is no real solution but the complex solution can be found.[/*][/list]

[br][color=#0000ff]Example 1. [color=#000000]Solve [math] 2x^2-10x+12=0.[/math] [/color][/color][br][br]The equation is in a standard form, so [color=#0000ff][color=#000000][math] a=2,\, b=-10 [/math][/color][/color]  and [color=#0000ff][color=#000000][math] c=12.[/math][/color][/color] By substituting these to the formula, we get the solutions:[br][br][math] \begin{array}{rcl}[br]x&=&\frac{-(-10)\pm\sqrt{(-10)^2-4\cdot 2\cdot(12)}}{2\cdot 2}\\[br]&=&\frac{10\pm \sqrt{100-96}}{4}\\[br]&=&\frac{10\pm \sqrt{4}}{4}\\[br]&=&\frac{10\pm 2}{4}\\[br]\end{array}[/math][br][br][math] x=\frac{10+2}{4}=3 \;\; \vee (=\text{or})\;\; x=\frac{10-2}{4}\approx 2 [/math][br][br][color=#0000ff]Example 2. Solve equation [math] -3x^2+5x=0.[/math] [/color] [br][br]The equation is in a standard form, so [math] a=-3,\, b=5 [/math]  and [math] c=0 [/math], because constant is missing from the original equation. By substituting these to the formula, we get the solutions:[br][br][math] \begin{array}{rcl}[br]x&=&\frac{-5\pm\sqrt{5^2-4\cdot (-3) \cdot 0}}{2\cdot (-3)}\\[br]&=&\frac{-5\pm \sqrt{25+0}}{-6}\\[br]&=&\frac{-5\pm \sqrt{25}}{-6}\\[br]&=&\frac{-5\pm 5}{-6}\\[br]\end{array}[/math][br][br][math] x=\frac{-5+5}{-6} =\frac{0}{6} = 0 \;\; \vee (=\text{ or })\;\; x=\frac{-5-5}{-6}=\frac{-10}{-6}=\frac{5}{3}. [/math][br][br][color=#0000ff]Example 3. [/color] Solve equation [math] 2x^2-8=0.[/math] [br][br]The equation is in a standard form, so [math] a=2,\, b= 0 [/math] (linear term is missing)  and [math] c=-8.[/math] By substituting these to the formula, we get the solutions:[br][br][math] \begin{array}{rcl}[br]x&=&\frac{0\pm\sqrt{0^2-4\cdot 2\cdot 8}}{2\cdot 2}\\[br]&=&\frac{\pm \sqrt{0+64}}{4}\\[br]&=&\frac{\pm \sqrt{64}}{4}\\[br]&=&\frac{\pm 8}{4}\\[br]&=&\pm 2[br]\end{array}[/math][br][br][math] x=-2 \;\; \vee (=\text{ tai })\;\; x=2 [/math][br][br][br][color=#0000ff]Example 4. [/color] Solve [math] x(2x-3)-3x(1-x)=-1.[/math] [br][br]Let us simplify the equation first by removing brackets and combining like terms:[br][br][math]\begin{array}{rcll}[br]x(2x-3)-3x(1-x)&=&-1\\[br]2x^2-3x-3x+3x^2&=&-1\\[br]5x^2-6x&=&-1\\[br]5x^2-6x+1&=&0&|\text{Standard form!}[br]\end{array}[br][/math][br][br]As the equations is of degree two (the highest power of the variable), it must be given in the standard form. [color=#0000ff]Values of parameters in a formula are always looked from a standard form.[/color][br][br][math] a=5,\; b=-6,\; c=1[/math][br][br][math]x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 5\cdot 1}}{2\cdot 5}=\frac{6\pm 4}{10}\\ \vspace{15mm}[br] x=\frac{6+4}{10}=1 \;\;\text{or}\;\; x=\frac{6-4}{10}=\frac{2}{10}=\frac{1}{5}[/math][br] [br]