solve for x: 6x^2+7x-3?

6x^2+7x-3
(2 x+3) (3 x-1)=0
2x+3=0 3x-1=0
2x=-3 3x=1
x=-3/2 x=1/3

x=-3/2,1/3 answer//

Factorise: (3x-1)(2x+3)=0
x=1/3 or -3/2

The discriminant is the fee of the expression below the unconventional sign interior the quadratic formula it particularly is b^2-(4ac) on the grounds which you're coping with a quadratic equation interior the variety ax^2+bx+c=0 you will locate the discriminant. subsequently a=5/6, b=-7 and c=-6/5. Plug the numbers into the discriminant, it particularly is b^2-(4ac) as above, to locate the fee of the discriminant. (i will go away that as much as you on the grounds which you're in a variety larger that Algebra a million.......) If the discriminant you detect is: detrimental - the concepts are complicated (there are no longer any actual concepts) corresponding to 0 - there is one actual answer powerful - there are 2 actual concepts.

1/3

6x^2 + 7x - 3
= 6x^2 + 9x - 2x - 3
= (6x^2 + 9x) - (2x + 3)
= 3x(2x + 3) - 1(2x + 3)
= (2x + 3)(3x - 1)

It is a hard question for me, sorry.

I am putting the expression equal to 0. Hope this is correct.

Using ac method

a = 6
c = -3

a times c = -18
Factors of -18 to give +7 (the middle number) when added = +9 - 2

Rearranging gives

6x*2 +9x - 2x -3 = 0

Bracketing

3x[2x + 3] - 1[2x + 3] = 0

Take out the common bracket gives

[2x + 3][3x - 1] = 0

Either 2x + 3 = 0 and x = -3/2

Or 3x -1 = 0 and x = 1/3

[If an answer is zero when 2 numbers are multiplied together one of the numbers must be equal to zero]

to find the roots plug the following coefficients into the quadratic formula :
a = 6
b = 7
c = -3


x = {-7 +/- [7^2 -4(6)(-3)]^1/2 } / 2(6)
x = {-7 +/- [49 + 72]^1/2 } / 12
x = {-7 +/- 11} / 12
x = 0.33333 , -1.5

6x^2+7x-3=0 - equate to zero

(3x - 1)(2x + 3)=0 - factor out

(3x - 1)=0 - equate each to zero
(2x + 3)=0

x = 1/3 , x = -3/2 - solve for x in each factor

x= -6.39