Question

ax^2+bs+c=0

Answer 1

isnt it ax^2+bx+c=0

Answer 2

You start out with the quadratic:
ax² + bx + c = 0

You want to divide by 'a'
x² + (b/a)x + c/a = 0

Put the constant on the other side
x² + (b/a)x = −c/a

Complete the square...
x² + (b/a)x + (b/2a)² = −c/a + (b/2a)²

Factor the left side:
(x + (b/2a))² = −c/a + (b/2a)²

Simplify:
(x + (b/2a))² = (b/2a)² − c/a

(x + (b/2a))² = b²/(4a²) − c/a

(x + (b/2a))² = b²/(4a²) − (4ac)/(4a²)

(x + (b/2a))² = (b² − 4ac) / (4a²)

Square root to isolate 'x'
x + (b/2a) = ±√[ (b² − 4ac) / (4a²) ]

x + (b/2a) = ±√[ b² − 4ac ] / (2a)

x = −(b/2a) ±√[ b² − 4ac ] / (2a)

x = [ −b ±√( b² − 4ac ) ] / (2a)