Solve using the quadratic formula: x2 – 3x – 3 = -5?

x^2 - 3x - 3 = -5

x^2 - 3x + 2 = 0

(x - 2)(x - 1) = 0

x = 2 and 1

The quadratic formula is -b+or-[b^2-4ac]/2a, and the abc are the coefficients from your equation when in this form: ax^2+bx+c. So using that information, start substituting numbers into the formula:

-(-3)+or-[(-3)^2-4(1)(2)]/2(1) and do the math

3+or-[9-8]/2

3+or-(1)/2

3+or-0.5

3+0.5=3.5 or

3-0.5=2.5

x=3.5 or x=2.5

Now you have to plug these values into the equation to see which is correct.

I didn't see your -5, so my math is wrong. You have to add 5 to both sides, making c=+2. I went back and fixed it.

wpf.

Hi , I think you the formula and you shall get the answer.Let's try to derive an formula for a general quadratic terms . Let us take a quadratic equation ax^2+bx+c = 0 Now we have to derive the value of "x". If p*q = 0 then we get EITHER p = 0 OR q = 0. So if I get in terms of p,q we get our answer.So let us try. ------> Multiply by 4a on both sides we get (2ax)^2+2(2ax)(b)+4ac = 0. ------->Now let us try to get in the formula of (a+b)^2.Addind and subtracting b^2 on the left hand side we get (2ax)^2+2(2ax)(b)+b^2--b^2+4ac = 0 --------> Now I write the first three terms as ( 2ax+b)^2 and take "--" common in the second so I get a formula of p^2--q^2 that can be written in the form of (p+q)(p--q) ---------> ( 2ax+b)^2--{sqrt(b^2--4ac)}^2 = 0 ---------> {2ax+b+sqrt(b^2--4ac)} {2ax+b--sqrt(b^2--4ac)} = 0 ----------> x = [--b+(or)--{sqrt(b^2--4ac)}]/2a NOTE:This formula is applicable whan quadratic equation is in standard form. Now let us solve your problem taking a = 1 b = --3 c = --8 (By transforming 5 on the other side) Apply the above formula x = [--(--3)+(or)--{sqrt((--3)^2--4(1)(--8)}... By solving you get x = [3+(or)--sqrt41]/2 ANSWER: x = [3+(or)--sqrt41]/2

X2-3X=-2

X2-X=-2/3

X=-2/3

r u n lyke 7 or 8 th grade

pre algebra

add +3 to -3 and -5

3x crosess out

u have x2-3x= -8

add 3x to both sides

u have x2=IDK

By quadratic formula, do you mean: x=-b [+ or -] square root of [b^2 - 4ac] all divided by [2a]?

Well then, a=1 b=-3 and c=-3

Plug it in...

But I don't know what you use the -5 for...

Add like terms together making -1x -3 = -5. Then you get rid of what is added to the variable and coefficent on both sides of the equation so you then -1x=-5. Divide both sides of the equation by -1 and you get

x=5

x² - 3x - 3 = - 5

x² - 3x - 3 + 5

x² - 3x + 2 = 0

(a = 1, b = - 3, c = 2)

Solution set: x{1, 2}.

See solution below:

http://i533.photobucket.com/albums/ee339/revo_emag...

Put it in the quadratic formula.

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

x = 2 and/or 1