Quadratic formulas?

Question

how do I do quadratic formulas for ex: 3x^2-4x-5=0 I don't understand how to get the two problems in the parenthasis

Answer 1

The Quadratic Formula is: x = [-b +/- sqrt(b^2 - 4ac)] / 2a

a, b, and c represent the values of x^2, x, and the constant in your equation. So in your example, 3x^2 - 4x - 5 = 0, a would be 3, b would be -4 and c would be -5.

You would plug in your these values into the Quadratic Formula like this:

x = [-(-4) +/- sqrt((-4)^2 - 4(3)(-5))] / 2(3)

Now you can solve as a regular equation:

x = [4 +/- sqrt(16 + 60)] / 6
x = [4 sqrt(76)] / 6, [4 - sqrt(76)] / 6

I'm not sure how far you need to go but you can take it from here I'm sure. Hope this helps. Remember You will have TWO answers for this problem because you take the positive square root and the negative square root separately. Good Luck.

Check here for more help: http://www.purplemath.com/modules/quadform.htm

Answer 2

The quadrtic formula is
y= -b + or - the sqrt of b -4(a)(c)
All over 2(a)

So in this case, plug 3 into all (a) slots, -4 into all (b) slots, and -5 into all (c) slots.

You would get
4 plus or minus the sqrt of -4 - 4(3)(-5)
All over 2(3)

Now, just follow the order of operations to get the asnwer. Due to the + or - sign at the begining, you should get two answers. Plug each one into x to find out the proper solution. Anything that is pluged into x, should make the entire equation equal zero if it is the correct solution.

Answer 3

the quadratic formulation is used to remedy the conceivable variable recommendations in quadratic equations, that are equations wherein the a variable is squared. the formulation itself is: x = -b (+/-) (sq. root of (b^2-4ac) / 2a so given 2x^2 +9x + 4 a= 2 b= 9 c= 4 so then plugging it into the equation: x= (-9) (+/-) sq. root (9^2-4(2) (4) / 2(2) simplified: x= -9 (+/-) sq. root (40 9) / 4 x= (-9 + 7)/ 4 or (-9 - 7) /4 solved: x = -a million/2, or -4 desire that enables! solid success...

Answer 4

the formula is y= -b+- the square root of b^2-4ac all over 2a.
where a is the x^2 number in ur equation and b is the x coefficient and c is the last number

so a is 3 b is -4 and c is -5

Answer 5

3x^2-4x-5=0 has the form
ax^2+bx+c=0 where
a=3
b=-4 &
c=-5

the quadratic formula is x=(-b+/-√(b^2-4ac))/2a
substituting
x=(4+/-√(16+4*3*5))/6
x=(4+/-√76)/6
x=2/3+/-(1/3)√19
x=(2+√19)/3, (2-√19)/3

Answer 6

Here's the quadratic formula:
x=-b±√b^2-4ac/2a
a=3,b=-4,c=-5
x=(-(-4)±√(-4)^2-4(3*-5))/2(3)
x=(4±√16-4(-15))/6
x=(4±√16+60)/6
x=(4±√76)/6
x=(4±√4*19)/6
x=(4±2√19)/6
x=(2±1√19)/3

I hope this helps!

Answer 7

3x²-4x-5=0
D = -4² - 4.3.-5
d = 16 + 60
d = 76
x = (4 +/- \/76) : 2.3
x' = (4 + 2\/19) : 6
x' = 2/3 + \/19 over 3
x" = (4 - 2\/19) : 6
x' = 2/3 - \/19 over 3
Solution: {x element of R | x' =2/3 + \/19 over 3 and x' = 2/3 - \/19 over 3}
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Answer 8

The two roots of this equation are irrational.

Answer 9

F.O.I.L. First, outside, inside, last, in that order and then there you'll have your answer.

Answer 10

x= [-b +- square root ( b^2 - 4ac)] / 2a
a= 3
b= -4
c= -5

x= (-4 +- 8.71)/ 6
x= -2.1 or x= .785
(x+2.1) (x-.785)