solve ? ax2+bx +c?

It is not an equation we can just find "zeros" for this polynomial.

in mathematics "zero" is a value which if substituted in a polynomial, in place of a variable (here x) makes its value equal to 0.

the two "zeros" of this polynomial are

{ - b - square root of "b2-4ac" } / 2a

and { -b + square root of "b2-4ac" } / 2a

But if this is an eqution viz. ax2+bx+c = 0

then we can solve it

in mathematics "solution" or "answer" is a value which if substituted in an eqution, inplace of variable (here x) makes its two sides equal

then two solutions of given equation can be obtained be placing each "zero" equal to 0 separately

please answer my questions. i shall be very thankful to you.

If this expession is=0,it is a quadatic equation

Solution is

x=minus b plus or minus SQUARE ROOT OF( b^2-4ac)/2a

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ax^2 + bx + c = 0

=> a^2x^2 + abx + ac = 0

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

=> (ax + b/2)^2 = (b^2 -- 4ac)/4

then ax + b/2 = ±1/2√(b^2 -- 4ac)

=> ax = [-- b/2 ±1/2√(b^2 -- 4ac)]

= x = [-- b ±√(b^2 -- 4ac)] / 2a

TELL ME THE VARIABLES TO SOLVE

if you solve for x, it is the quadratic formula.

x1= (-b + sqrt(b^2 - (4*a*c)))/(2*a)

x2= (-b - sqrt(b^2 - (4*a*c)))/(2*a)

it has two results.

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

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

x = -b ± √ (b² - 4ac) / 2a

x = -b ± √ (b² - 4ac) / 2a