2007-01-18 12:28:41 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
the real base formula is: ax^2 + bx + c = 0
2007-01-18 12:36:53 · answer #1 · answered by Jake 2 · 1⤊ 0⤋
the quadratic formula is the formula to get the roots or solutions to the equation
here it is:
x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
where a,b,c correspond to:
ax^2 + bx + c = 0
2007-01-18 20:35:07 · answer #2 · answered by AibohphobiA 4 · 2⤊ 0⤋
If ax^2 +bx +c = 0
then x = [-b +/- sqrt(b^2-4ac)] /2a
Just plug the values of a,b, and c into the above formula and you will get the twoo roots of the equation.
2007-01-18 20:35:34 · answer #3 · answered by ironduke8159 7 · 2⤊ 0⤋
The quadratic formula:
x=-b屉b^2-4ac/2a
for this kind of equation:
ax^2+bx+c=0
x^2+8x+16=0
I hope this helps!
2007-01-18 20:34:06 · answer #4 · answered by Anonymous · 0⤊ 0⤋
ax2+bx+c
hm ax2 is supposed to be squared. The quadratic formula is to get the solution to the equation
2007-01-18 20:31:51 · answer #5 · answered by jon88nhan 1 · 0⤊ 0⤋
thats like an equation with a square in it...
x^2+2x-8
which can be factorised into
(x+4)(x-2)
These are used to make graphs with a curve in so in this example the graph would cross the y axis at -4 and 2 but u probably dont need this lot
2007-01-18 20:32:57 · answer #6 · answered by Anonymous · 0⤊ 0⤋
[ -b +,- sqrt(b^2-4ac) ] / 2a
where A is the coeffecient (number in front of) x^2
b is the coeffecient of x
and c is the constant in a quadratic equation
2007-01-18 20:33:28 · answer #7 · answered by brothergoosetg 4 · 2⤊ 0⤋
-b+ or - the square root of b squared - 4ac all divided by 2a
2007-01-18 20:33:13 · answer #8 · answered by packerswes4 5 · 1⤊ 0⤋
(-b ± â(b²-4ac)) / (2a)
2007-01-18 20:33:03 · answer #9 · answered by 7 · 2⤊ 0⤋