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^2-7x-5=0.[/math] [/color][/color][br][br]The equations is in a standard form, so [color=#0000ff][color=#000000][math] a=2,\, b=-7 [/math][/color][/color]  and [color=#0000ff][color=#000000][math] c=-5.[/math][/color][/color] By substituting these to the formula, we get the solutions:[br][br][math] \begin{array}{rcl}[br]x&=&\frac{-(-7)\pm\sqrt{(-7)^2-4\cdot 2\cdot(-5)}}{2\cdot 2}\\[br]&=&\frac{7\pm \sqrt{49+40}}{4}\\[br]&=&\frac{7\pm \sqrt{89}}{4}[br]\end{array}[/math][br][br][math] x=\frac{7+\sqrt{89}}{4}\approx 4.1 \;\; \vee (=\text{or})\;\; x=\frac{7-\sqrt{89}}{4}\approx -0.6 [/math][br][br][br][color=#0000ff]Example 2.  [color=#000000]Solve [math] x(2x-3)-3x(1-x)=-1.[/math] [/color][/color][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]