quick math problem - quadratic-like equations?

2x^2 + x - 15 = 0
factoring

(2x -5)(x +3) = 0

2x - 5 = 0
x = 5/2

x+3 = 0
x = -3

x = 5/2 and x = -3

As they are able to't be factorised, you should apply the quadratic equation: x = {-b +or- sqrt(b^2-4ac)}/2a the place a is the coefficient of x^2 (the huge style infront of x^2), b is the coefficient of x (huge style infront of x) and c is the consistent. For a million), a=3, b= -2, c=4 (that is important to get the signs and indicators precise!), and as such: x = {2 +or- sqrt(4 - 40 8)}/6 ~ we can't take the sq. root of a destructive huge style, so there is not any real answer to this situation. 2) {-5 +or- sqrt(25-(-4))}/2 ~ this would the two settle for in terms of sqrt(29), or worked out with a calculator, yet there is not any 'advantageous' answer to it. Please observe that the 1st person's reaction on your question would provide you advantageous solutions, yet is misguided. they have taken a worry-loose element to get: x(3x-2) = -4, and then used the tactic you employ to unravel an equation while the equation is comparable to 0. The equation it is not equivalent to 0, so assuming any area of that's comparable to 0 is misguided. in case you have any doubts, replace the values into the unique equations, and it will tutor that those are incorrect solutions.

(2x - 5)(x + 3) = 0
x = 5/2 , x = - 3
OR
x = [ - b ± √ (b ² - 4 a c ) ] / 2 a
x = [ - 1 ± √ (1 + 120 ) ] / 4
x = [ - 1 ± 11 ] / 4
x = 10/4 , x = - 12/4
x = 5/2 , x = - 3

2X^2 + X -15 = 0
2*15=30 First Multiply A and C.
6*5=30 6-5=1 Find the two factors of the product that add * to equal B.
(X + 6)(X - 5) Plug in the factors that worked.
(X + 6/2)(X - 5/2) Divide by A.
(X + 3) (X - 5/2) Reduce to lowest terms.
(X + 3) (2X - 5) Move any denominators left to the front of *** the variable in the same parentheses.

The problem is now solved. I hop this helped.

i THINK i hav it:
use the factor diamond method [i learned it in school]
and you get
2x^2 +6x-5x-15=0
than
2x(x+3)-5(x+3)=0
and than group again
(2x-5)(x+3)=0
if you need to find values for x. . .
than set each group as being =0 using the 0 property. . .
2x-5=0, so x=5/2
and
x+3=0, so x=-3
. . . and there you go


not bad for a soon-to-be freshman?

well what you have to do is you factored wrong it would be:
(2x)(2x) than you would have to FOIL(first,inner,outter,last) and solve it out from there or the other way you do it is you dont even have to factor you can just use the quadratic formula:
A=2
B=1
C=15
than you just apply it into your quadratic formula and solve it from there((this is probably the eaiser way))

easiest way is to graph the problem, then see the answers (where the graph crosses the x-axis).

Here are the results
(2x-5)(x+3)

another way is to plug into quadratic formula, get 5/2 and -3, and come up with factors from there

quadratic equation is -b (plus or minus) Root (bsquared minus 4ac) ALL over 2a

2x.x and (-5).3

because 2x.(3) + x.(-5) = +x

so (2x-5)(x+3) the result

2x^2 + x - 15 = 0
2x^2 + 6x - 5x - 15 = 0
(2x^2 + 6x) - (5x + 15) = 0
2x(x + 3) - 5(x + 3) = 0
(x + 3)(2x - 5) = 0

x + 3 = 0
x = -3

2x - 5 = 0
2x = 5
x = 5/2 (2.5)

∴ x = -3 , 5/2 (2.5)