Can anyone answer this one? 3x^2=7x+20?

3x² - 7x - 20 = 0
(3x + 5)(x - 4) = 0
x = - 5/3 , x = 4

3x^2 = 7x + 20
3x^2 - 7x - 20 = 0
(3x + 5)(x - 4) = 0

3x + 5 = 0
3x = -5
x = -5/3
x = -1 2/3

x - 4 = 0
x = 4

∴ x = -1 2/3 , 4

3x^2=7x+20
3x^2-7x=20
3x^2-7x-20=0
3x^2-(12x-5x)-20=0
3x^2-12x+5x-20=0
3x(x-4)+5(x-4)=0
(3x+5)(x-4)=0
Either (3x+5) or (x-4) should be 0.

If 3x+5=0;
3x=-5;
x=-5/3

If x-4=0;
x=4

3x^2 = 7x + 20

Subtract 7x and 20 from both sides
3x^2 - 7x - 20 = 0

Factor
(3x + 5)(x - 4) = 0

Set both factors equal to 0 and solve forx
3x + 5 = 0
x = -5/3

x - 4 = 0
x = 4

This can be factored into (3x + 5)(x - 4) = 0. Since either of these factors can be 0, then we set each one equal to 0 and solve for x in each:

3x + 5 = 0
3x = -5
x = -5/3

or

x - 4 = 0
x = 4

To see if they are roots of the equation, plug each one in and see if their functional value is 0. If they are, then we have the correct solutions. I have checked both of them, and they both work.

this kind of sum is solved by middle term factor and the procedure is as follows:

3x square=7x+20
or 3x square -7x -20=0
or 3x square - (12-5)x -20=0
or 3x square -12x +5x -20 =0
or 3x ( x - 4) + 5 ( x -4 ) =0
or (x-4) (3x +5) =0

either,
x - 4=0
or x=4

OR

3x+5=0
or 3x=5
or x= 5/3.

answer : x=4
or x=5/3

hope you can follow the procedure.

(3x+5)(x-4)=0