Question

3(x2 – 13x +41) +8=11

Answer 1

Start with

`3(x^2-13x+41)+8=11`

Subtract 8 from both sides:

`3(x^2-13x+41)=3`

Divide both sides by 3:

`x^2-13x+41=1`

Subtract 1 from both sides:

`x^2-13x+40=0`

Solve with the usual formula

`ax^2+bx+c=0\iff x = (-b\pm√(b^2-4ac))/(2a)`

to get

`x^2-13x+40=0\iff x = (13\pm√(9))/(2) = (13\pm 3)/(2)`

So, the two solutions are

`x_1 = (13+3)/(2)=8,\quad x_2 = (13-3)/(2)=5`