Verified answer

Factor it:

5x^2 - 25x - 30 = 0 (factor out a 5)

5(x^2 - 5x - 6) = 0 (factor by finding two integers that multiply to equal -6, but add to equal -5, which are 1 and -6)

5(x - 6)(x + 1) = 0 (because the equation equals 0, separate the parentheses into different equations)

x - 6 = 0

x = 6

x + 1 = 0

x = -1

ANSWER: x = 6, -1

FACTOR!!

(-30)(5)= (-60)

so the numbers have to have a product of (-60) and a sum of (-25)

In this instance (-30) and (5) will work quite nicely.

Now, rewrite the equation using those two numers as the middle term, like this:

5x^2 - 30x +5x - 30 = 0

now factor by grouping (put a parenthsese around the first two terms and the second two terms):

(5x^2 - 30x) (+5x - 30) = 0

now within each group, take out the common factor:

5x(x-6) +5(x-6)=0

now rewrite as:

(5x+5)(x-6)=0

set each factor to zero:

5x+5=0

x-6=0

now solve for x in each equation and you get:

x= (-1)

x= 6

5x^2 - 25x - 30 = 0

(5x^2 - 25x - 30)/5 = 0/5

x^2 - 5x - 6 = 0

x^2 + x - 6x - 6 = 0

(x^2 + x) - (6x + 6) = 0

x(x + 1) - 6(x + 1) = 0

(x + 1)(x - 6) = 0

x + 1 = 0

x = -1

x - 6 = 0

x = 6

∴ x = -1 , 6

By using factorization of that equation in that way :

( 5x + 5)( x - 6 ) = 0

so,

(5x + 5) = 0 or (x - 6) = 0

5x = -5 or x=6

x= -5/5 or x=6

x= -1 or x=6

The answer is x = { 6, -1 } ........... 6 or -1

bye :)

5x² - 25x - 30 = 0? ----> first thing, divide both side by 5

x² - 5x - 6 = 0?

(x - 6)(x + 1) = 0

x - 6 = 0 -------> change sign on -6, then move to other side

x = 6

x + 1 = 0 -------> change sign on 1, then move to other side

x = -1

Therefore x = 6 or x = -1

Certainly!

Notice that the sum of 5 and 25 is 30-- so what would make them add up to 30?

Well, somehow you've got to cancel that minus sign in front of 25x...

*lightbulb!*

Notice that x is squared in the a term (where 5 is the coefficient)... that means you can use a negative number to cancel the minus sign in front of 25x without setting 5x² to a negative!

And what negative number would you have to multiply by those two terms to have them add up to 30?

;-)

5x² - 25x - 30 = 0

5x² - 30x+5x - 30 = 0

5x(x-6)+5(x-6)=0

(x-6)(5x+5)=0

so either x-6=0 ==> x=6

or 5x+5=0 ==> x=-1

or apply

x=(-b+ (sqrt)(b²-4ac))/2a

or

x=(-b- (sqrt)(b²-4ac))/2a

a> coff of xsq

b coff of x

c=constant term

(5x +5 )(x -6 )

if u dont know how i did that use the quadratic formula

which is [-b+/- sqrt( b^2-4ac)]/2a

ax^2+bx+c=0

so a=5 b=-25 c=-30

x=-1 and 6

Consider the above equation as (ax^2)+bx+c=0

then substitute in the equation,

x= ((-b) +(b^2-4ac)^(1/2))/2a

x=((-b) -(b^2-4ac)^(1/2))/2a

5(x^2-5x-6)=0

5(x-6)(x+1)=0

x-6=0 or x+1=0

x=6 or x= -1