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

$ax^2+bx+c=0\iff x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

to get

$x^2-13x+40=0\iff x = \dfrac{13\pm\sqrt{9}}{2} = \dfrac{13\pm 3}{2}$

So, the two solutions are

$x_1 = \dfrac{13+3}{2}=8,\quad x_2 = \dfrac{13-3}{2}=5$