What are the roots of the polynomial 3x^2=11x+4?

Solution:

3x^2 - 11x - 4 = 0
3x^2 -12x +x -4 =0
3x(x -4) +(x-4) =0
Thus,
(3x + 1)(x - 4) = 0

x = -1/3 or x - 4

x =4 or x = -(1/3)

3x² - 11x - 4 = 0
(3x + 1)(x - 4) = 0
x = - 1/3 , x = 4

3x^2 - 11x - 4 = 0

3x^2 - 12x + x - 4 = 0

3x(x - 4) +1(x - 4) = 0

(x - 4)(3x + 1) = 0

x - 4 = 0 or 3x + 1 = 0
x = 4 or x = -1/3

The 2nd line is a trick for these sort of quadratics when the coefficient of x^2 is more than 1.
Multiply coef of x^2 (3) by constant (-4) to get -12.
Find 2 numbers which multiply to make -12 but add to -11 (the coef of x).
These are -12 & +1.
Then rewrite the middle term -11x as -12x + x (using these 2 numbers).
You can now find the common factor of 1st 2 term, and common factor of 2nd 2 terms. They will have the same bracket in common, i.e. a common factor and it can be fully factorised into 2 brackets.
Then solve as normal for quadratics with left bracket = 0
or right bracket = 0.

Practice, this always works with this sort.

Good luck.

3x^2 = 11x + 4
3x^2 - 11x - 4 = 0
(3x + 1)(x - 4) = 0

3x + 1 = 0
3x = -1
x = -1/3

x - 4 = 0
x = 4

∴ x = -1/3 , 4

3x^2=11x+4
3x^2-11x-4=0
(3x+1)(x-4)=0
For (3x+1)(x-4) to equal zero, either 3x+1=0, or x-4=0
If 3x+1=0, 3x= -1 and x= -1/3
If x-4=0, x=4
The roots are 4 and -1/3

3x^2=11x+4

3x² - 11x - 4 = 0

(3x + 1) (x - 4) = 0

3x + 1 = 0
x = -1/3

x - 4 = 0
x = 4

The roots are -1/3 and 4.

rearrange the equation:
3x^2 - 11x - 4 = 0

factorize:
(3x + 1)(x - 4) = 0

so, the roots are
x = -1/3 and 4

3x^2 - 11x - 4 = 0

(3x + 1)(x - 4) = 0

x = -1/3 or x - 4