Question

solve ax2+bx+c=0
solve 3x2-12x+27=0

Answer 1

solve as in find the roots?

Answer 2

\[ax^{2} + bx + c = 0\\a \left[ x ^{2} + \left( \frac{ b }{ a } \right)x \right] = -c\\a \left[ x ^{2} + \left( \frac{ b }{ a } \right)x +\frac{ b ^{2} }{ 4a ^{2} } \right] = \frac{ b ^{2} }{ 4a } -c\\x ^{2} + \left( \frac{ b }{ a } \right)x +\frac{ b ^{2} }{ 4a ^{2} } = \frac{ b ^{2} }{ 4a ^{2} } -\frac{ c }{ a }\\\left( x + \frac{ b }{ 2a } \right)^{2} = \frac{ b ^{2} - 4ac }{ 4a ^{2} }\\x + \frac{ b }{ 2a } = \frac{ \pm \sqrt{b ^{2} - 4ac } }{ 2a }\\x = \frac{ -b \pm \sqrt{b ^{2} - 4ac } }{ 2a }\]

Answer 3

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)

Answer 4

So, now that you have a solution for the general equation, you don't have to go back and "re-create the wheel". You can just use the end result, the quadratic formula, to substitute for those "a", "b", and "c". That second question has those lined up for you. So, just substitute and solve for the second question.

Answer 5

All good now, @4everdream322 ?