(5x-4)(2x-3)=17?

Question

I have to use the foil method to solve it and I know how to multiply it but what do I do after that

Answer 1

First: 10x^2 +
Outer: -15x +
Inner: -8x +
Last: 12

So this sums up to: 10x^2 - 15x - 8x + 12
Which reduces to: 10x^2 - 23x + 12
Then you subract 17: 10x^2 - 23x - 5

Now you can use the quadratic formula:
x =[ -b +/- sqrt(b^2 - 4ac)]/2a

x = [23 +/- sqrt(23^2 - 4*10*-5)]/(2*10)

x= [23 +/- sqrt(529+200)]/20

x=[23 +/- sqrt(729)]/20

x=[23 +/- 27]/20

x= 50/20 or -4/20

x= 5/2 or -1/5 (in decimals: x=2.5 or -0.2)

Answer 2

I think that some were making this too confusing.

Multiply it out and you get 10xsquared - 15x -8x + 12 = 17
Collect like terms 10xsquared - 23x - 5 = 0
To find the factors of this equation multiply the first and last terms...so the 10 and the 5. That gives you 50. What 2 terms can you find that would both multiply to equal 50 and add up to equal 23. Very simply 25 and -2. (25 x 2 = 50; 25 - 2 = 23)
Substitute those numbers in for the middle term.
(10xsquared + 2x) - (25x - 5) = 0
Remove the common factor from each bracket.
2x(5x + 1) - 5 (5x + 1) = 0
(2x - 5) (5x + 1) = 0

Now set each bracket to equal zero and solve.
2x - 5 = 0 2x = 5 x = 5/2
5x + 1 = 0 5x = -1 x = -1/5

Answer 3

(5x-4)(2x-3)=17

As you know, foil method, so you end up getting:

10x^2-8x+12=17

Subtract 12 from both sides in order to combine like fragments

10x^2-8x=5

Now, defoil the left side by finding similarities

2x(5x-4)=5

Solve for both x's

2x=5

Divide by 2, and get x=5/2

5x-4=5

Add four to both sides, cancelling out the -4 on the left

5x=1

Divide by 5, and you get x= 1/5

Answer 4

Cant follow foil method because one side has to be zero. I mean, you could find X that gives you 17 for the left side to equal the right side but I cant think of any combonations that give you positive 17. Id follow the suggestion of other people - quadratic.

Answer 5

It opens down simply by fact the x^2 term has a unfavorable coefficient (-.05) They improve until eventually they hit a top and the drop dow to 0 over the 2d a million/2 of the revenues era they are going to be at a top. they're going to drop to 0 on day eighty 4 the top wiill take place on the (-40 two)/-2*-.05 = forty second day .-.5*40 two^2 +40 two^2 +23 = 905 top revenues Vertex at (40 two,905) top on 2 in view that 40 two^2-4(.5)(23) is constructive The strategies indicat the style of days over which tickets have been offered

Answer 6

use foil and you get 10x^2 - 23x + 12 = 17

subtract 17 and get 10x^2 - 23x - 5 = 0

now use the quadratic formula to solve.

Quadratic formula is using the variables a, b, and c.
In this problem,
a = 10
b = -23
c = -5

0 = (-b + square root(b^2-4*a*c)/2a
0 = (-b - square root(b^2-4*a*c)/2a

that will give you your two answers for the zeros

Answer 7

(5x - 4)(2x - 3) = 17
10x^2 - 15x - 8x + 12 = 17
10x^2 - 23x + 12 = 17
10x^2 - 23x - 5 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-23) ± sqrt((-23)^2 - 4(10)(-5)))/(2(10))
x = (23 ± sqrt(520 + 200))/20
x = (23 ± sqrt(729))/20
x = (23 ± 27)/20
x = (50/20) or (-4/20)
x = (5/2) or (-1/5)

ANS : x = (5/2) or (-1/5)

Answer 8

use the above method you should get 2.5 and -0.2
good luck