Qudratic Formula?

First, factor out a five: 5(x^2+2x+4)=0 Now factor what is in the parentheses. In your equation, the 10x or whatever number is in the middle, is the sum of the products of the outer numbers and the inner numbers. Remember FOIL? First Outer Inner Last. So lets FOIL whats in parentheses. What can we multiply to get x^2? x *x (x +_ ) (x +_ ) what can we multiply to get 4, and add to get 2? Our signs are + and + so in our factored pair we have to have + and +. There is no two numbers that satisfy this equation so it can not be factored with this method. But hopefully the FOIL thing jogged your memory.

a) 4x² - 0x - 121 = 0 (2x + 11)(2x - 11) = 0 x = -5/2 or 5/2. b) 2x² - 3x - 2 = 0. (2x + a million)(x - 2) = 0 x = -a million/2 or 2. c) 15x² - 165x - 630 = 0 x² - 11x - 40 two = 0 (x + 3)(x - 14) = 0 x = -3 or 14. d) 36x² - 60x + 25 = 0. (6x - 5)(6x - 5) = 0. x = 5/6. e) 6x² + 11x - 10 = 0. (3x - 2)(2x + 5) = 0. x = 2/3 or -5/2.

5x^2 + 10x + 20 = 0
(5x^2 + 10x + 20)/5 = 0/5
x^2 + 2x + 4 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = 2
c = 4

x = [-2 ±√(4 - 16)]/2
x = [-2 ±√-12]/2 (imaginary number)
(no real roots)

First of all this problem is not solved using the quadratic equation. You had the right idea by factoring but you can simplify it first.

5x² + 10x + 20 = 0

Divide out 5...
x² + 2x + 4 = 0

Now factor (remeber the special factors)

A² + B² = A² + 2AB + B²
A² - B² = A² - 2AB + B²

x² + 2x + 4 = 0
x² + 2x + 2² = 0
(x + 2) ( x + 2) = 0

(x + 2)² = 0

x = - 2

the quadratic formula would have been much harder

For
ax² + bx + c = 0

x =

- b ± √ (b² - 4 a c)
------------------------
2a

To start off, what you're doing is actually factoring. The quadratic formula is: -b +- the square root of: b^2 + -4ac. And then all that divided by 2. Secondly, if you were to factor the equation you have, like what you did, you factored it wrong. (x + 10)(x+2)=x^2 + 12x + 20. Not 5x^2 + 10x + 20. Some equations just cannot be factored, so you would use the quadratic formula in that instance. Follow the formula and you get your two answers.

You were asking what happens to the middle number. When you factor, you always consider both the last, and middle number. You use the multiples of the second coefficient (bx), in this case 10x, and the numbers that add up to the last number, 20. You try to find a pair of numbers that works both ways, for both numbers. So what you do is list all the factors of the middle number (b) you can think of and make sure they add up to equal the sum of the last number (c). Take into consideration your first number (a), for if it's not just 1x, but something like 5x you will have to keep trying and make adjustments. Then check your factoring by multiplying it all out and you should get the same exact equation you started with. Then you solve for both x's.

Hope that's understandable!!

5x²+10x+20 = 0
x = -1±√3i

the equation you gave has no solution

Divide through by 5

x^2+2x+4 = 0

solve using [-b+-sqrt(b²-4ac)]/2a

[-2+-2sqrt(-3)]/2 = -1+2i*sqrt(3) or -1-2i*sqrt(3)

In this case you have two complex solutions

nope, just plug in the numbers to the quadratic formula.

when factorring out (x+10 and (x+ 5 u use FOIL which means multiply the first terms, multiply the 2 first terms and 2 last terms and add them, and then multiply the last terms. the middle number comes from the O and the I added together. U factored this wrong i dont think it has a rational solution. u have to use the quadratic formula look it up