Axe? +bx+c=0
x? +(b/a)x+c/a=0
x? +2×[b/(2a)]x+c/a=0
x? +2×[b/(2a)]x+[b/(2a)]? -[b/(2a)]? +c/a=0
x? +2×[b/(2a)]x+[b/(2a)]? =[b/(2a)]? -c/a
[x+b/(2a)]? =b? /(2a)? -4ac/(2a)?
[x+b/(2a)]? =(b? -4ac)/(2a)?
x+b/(2a)= √[(b? -4ac)/(2a)? ]
x+b/(2a)= [√(b? -4ac)]/(2a)
x=-b/(2a) [√(b? -4ac)]/(2a)
x=[-b √(b? -4ac)]/(2a)
Extended data:
Solution of quadratic equation in one variable;
First, the direct Kaiping method
The form is (x+a) 2 = b, when b is greater than or equal to 0, x+a= plus or minus radical number b, x=-a plus or minus radical number b; When b is less than 0. This equation has no real root.
Second, the matching method
1. The quadratic term is converted to 1.
2. Shift the term, with quadratic term and linear term on the left and constant term on the right.
3. Formula, add half the square of the first-order coefficient on both sides and change it into the form of (x = a) 2 = b.
4. Solve the equation by direct Kaiping method.
Third, the formula method
Now the equation is arranged in a general form: ax 2+bx+c = 0. Then substitute abc into the formula x = (-b √ (b 2-4ac))/2a, and (b 2-4ac is greater than or equal to 0).
Fourthly, factorization method.
If the algebraic expression on the left of the unary quadratic equation AX 2+BX+C = 0 is easy to decompose, then the factorization method is preferred.