*ax*x + bx +c = 0*

*this specific structure is known as a quadratic function, therefore x will be *

*`(-b+-4ac)/(2a)` *

`ax^2 + bx + c = 0`

Is a special type of quadratic equation. The x equals

-b (plus or minus) sqrt ((b^2 - 4ac)/2a)

ax² + bx + c = 0

devide the equation by a

x² + (b/a)x + c/a = 0

[x+(b/2a)]² -b²/4a² + c/a = 0

[x+(b/2a)]² - (b² - 4ac)/4a² = 0

[x+(b/2a)]² - [√(b² - 4ac)/√4a²]² =0

[x+(b/2a)]² - [√(b² - 4ac)/2a]² =0

{ x + b/2a + √(b² - 4ac)/2a }{ x + b/2a - √(b² - 4ac)/2a } = 0

{ x + [b+ √(b² - 4ac)]/2a }{ x + [b- √(b² - 4ac)]/2a } = 0

x = [-b - √(b² - 4ac)]/2a or x = [-b + √(b² - 4ac)]/2a

**x = [-b ± √(b² - 4ac)]/2a **

To solve ax*x+bx+c = 0.

Solution: We can write the equation like:

ax^2+bx+c= 0.....(1), as x*x = x^2. This is a second degree equation and is also called a quadratic equation which will be solved by writing the left as diffrence of an unknown square and a known square equal to zero:

LHS of (1) :

ax^2+bx+c = a[x^2+(b/a)x]+c

=a(x^2+(b/a)x+(b/(2a))^2] - b^2/(4a)+c. Here (b^2/(4a) is added and subtracted.

=a[x+(b/(2a))]^2 - (b^2-4ac)/4a .

=a{[x+b/(2a)]^2 - (b^2-4ac)/(2a)^2}

So the equation (1) now could be rewritten like:

a{[x+b/(2a)]^2 - (b^2-4ac)/(2a)^2} = 0. Or bt dividing by a which is not equal to zero, we get:

[x+b/(2a)]^2 - (b^2-4ac)/(2a)^2 = 0. Or

[x+b/(2a)]^2 = (b^-4ac)/(2a)^2. Taking the square root,

x+b/(2a) = +sqrt(b^2-4ac)/2a . Or x+b/(2a) = -sqrt(b^2-4ac)/2a. So we get 2 solutions from the se two results.

x1 = [-b+sqrt(b^2-4ac)]/(2a) Or

x2 = [-b-sqrt(b^2-4ac)]/(2a)