Answer:
[tex]x = \frac{ - b ± \sqrt{ {b }^{2} - 4ac } }{2a} [/tex]
Step-by-step explanation:
For example, we'll use this quadratic equation.
[tex] {x}^{2} + 5x + 6[/tex]
To understand how to plug it into the formula we need to know what each term represents.
[tex]a {x}^{2} + bx + c[/tex]
So the equation above would be put into the formula like this.
[tex]x = \frac{ - 5± \sqrt{ {5}^{2} - 4(1)(6) } }{2(1)} [/tex]
Then we would solve
[tex] \frac{ - 5± \sqrt{25 - 24} }{2} \\ \\ = \frac{ -5±1}{2} [/tex]
Now, the equation will branch off into one that solves when addition and one when subtraction.
[tex] \frac{ - 5 + 1}{2} = \frac{ - 4}{2} = - 2 \\ \\ \frac{ - 5 - 1}{2} = \frac{ - 6}{2} = - 3[/tex]
So x={-3, -2} (-3 and -2)
