How do i solve 11x^2=31x+6?

11x^2-31x-6=0
(x+2/11)(x-3)=0
x=3 x=-2/11

hi,
11x²=31x+6, 11x²-31x-6=0
lhs
11x²-33x+2x-6=0, 11x(x-3)+2(x-3)=0
(11x+2)(x-3) =0
11x+2= 0 , or x-3 = 0,
so x = -2/11 or x = 3

There will be two answer
x = - 2 / 11 or x = 3

11 x ² - 31 x - 6 = 0

x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a

x = [ 31 ± √ ( 961 + 264 ) ] / 22

x = [ 31 ± √ ( 1225 ) ] / 22

x = [ 31 ± 35 ] / 22

x = 66 / 22 , x = - 4/22

x = 3 , x = - 2 / 11

11x^2-31x-6=0
x=[31+&-(961+264)^1/2]/22=3 & -2/11

11x^2 = 31x + 6
11x^2 - 31x - 6 = 0
11x^2 + 2x - 33x - 6 = 0
(11x^2 + 2x) - (33x + 6) = 0
x(11x + 2) - 3(11x + 2) = 0
(11x + 2)(x - 3) = 0

11x + 2 = 0
11x = -2
x = -2/11

x - 3 = 0
x = 3

Hence, x = -2/11 or 3.

You try to get the equation into the form ax^2 +bx +c = 0. This case, that would be 11x^2 -31x -6=0. Then you use the quadratic equation. The equation is in my source link.

use the quadratic equation or factoring. hopefully you know what these are . . . =]

11x^2=31x+6
11x^2-31x--6=0
factorise it
(x-3)(11x+2) =0
therefore, either (x-3)=0 or (11x+2)=0
x-3=0
x=3
11x+2-0
11x=-2
x=-2/11

x=-2/11, 3

Hi,

11x² = 31x + 6

11x² - 31x - 6 = 0

(11x + 2)(x - 3) = 0

So,

x = -2/11, or 3

Hope this helps!