Question

(4x + 3)(x + 1) = 2

Explanation:

Step1: Expand left-hand side

$(4x+3)(x+1) = 4x^2 + 4x + 3x + 3 = 4x^2 + 7x + 3$

Step2: Rearrange to standard quadratic form

$4x^2 + 7x + 3 - 2 = 0$
$4x^2 + 7x + 1 = 0$

Step3: Apply quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=4, b=7, c=1$.
$x=\frac{-7\pm\sqrt{7^2-4\times4\times1}}{2\times4}=\frac{-7\pm\sqrt{49-16}}{8}=\frac{-7\pm\sqrt{33}}{8}$

Answer:

$x=\frac{-7+\sqrt{33}}{8}$ or $x=\frac{-7-\sqrt{33}}{8}$