This is a math algebra question! So if your a math wizard answer this question!?

Question

How do you derive -> ax(squared) +bx+c=0 to x= -b (posetive negative) /2a

Answer 1

i can do it i swear

ok

ax2+bx+c=0
ab2-b2+c=0
(a-b)b2=-c

there it's done. math whiz whooooo!

Answer 2

The technique is called "Completing the square"

ax^2+ bx + c = 0
subtract "c" from both sides of the equation
ax^2 + bx = -c

divide both sides of the equation by "a"


x^2 + (b/a)x = -(c/a)

to "complete the square" divide the (b/a) by 2 and then square
add this term to both sides of the equation

x^2 + (b/a)x + (b/2a)^2 = -(c/a) + (b/2a)^2

factor out the left side of the equation


( x + (b/2a)) ^ 2 = -(c/a) + (b/2a)^2

simplify the right hand fraction- the LCD is 4a^2

multiply the first fraction by (4a)/(4a) to get

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

add the fractions-putting the positive term first

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

take the square root of both sides of the equation-remembering that takin the square root produces a positive and negative root

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

subtract the (b/2a) from both sides of the equation
to get the standare formula

x = -b +/- b^2 - 4ac / 2a