How do you factorise -3x^2+2x+40?

-3x^2+2x+40 = 0
3x^2-2x-40 = 0
3x^2-2x+(1/3)-40-(1/3) = 0
{(sqrt3)x - (1/sqrt3)}^2 = 121/3
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Or by formula
x = {-b +/- sqrt(b^2 - 4ac) } / 2a

Here a = -3
b = 2
c = 40
Thus x = -10/3 or 4 .

-3x^2+2x+40
=40+2x-3x^2

Look here, 3 x40=120 & 12 x 10 =120 & also 12-10=2
so, we have:
40+(12-10)x -3x^2
=40+12x-10x-3x^2
=4(10+3x)-x(10+3x)
=(10+3x)(4-x)<===ANSWER

-3x^2+2x+40
= -3x^2 +12x - 10x + 40
= -3x(x - 4) - 10(x - 4)
=(x-4)*(-3x-10)
=(4-x)*(3x+10)

3x2-2x-4=0 6-2x-4=0(first multiply) 6-4=2x+0Then u shop # on one component and Letters on different 2/2=2x/2Then divide via the # you need to get journey of to purely be left with an x. x=a million !!!Remeber the substantial iskeep numbers on one component and the numbers that are linked to the letters on the different component. try this purely when you have executed the fundamental math like multiplying. The remeber...while u are waiting to chop up the numbers of the two component of the equivalent sign they alter into opposite. as an occasion, if the 4 replaced into +4 then once I circulate it over it rather is going to alter into -4. yet for that reason it replaced into -4 so it replace into +4.

To have two x values suggests an equation:-
-3x² + 2x + 40 = 0
3x² - 2x - 40 = 0
(3x + 10)(x - 4) = 0
x = - 10/3 , x = 4

--3x^2 + 2x + 40 = 0
or 3x^2 -- 2x -- 40 = 0
or 3x^2 -- 12x + 10x -- 40 = 0
or 3x(x -- 4) + 10(x -- 4) = 0
or (x -- 4)(3x + 10) = 0
either x = 4 or x = --10/3.

-3x^2 + 2x + 40
= -3x^2 + 12x - 10x + 40
= (-3x^2 + 12x) - (10x - 40)
= -3x(x - 4) - 10(x - 4)
= (-3x - 10)(x - 4)

-3x^2 +2x+40 = -3x^2 + 12x - 10x +40 =
= -3x(x - 4) -10(x-4)
= (-3x -10) (x-4)

(3x + 10) (-x + 4)

f
x=-2+-sqrt(4+480)/-6=-2+-22/-6
=4,-10/3
therefore to factorize
(x-4)(3x+10)