An algebra problem by Daehan Lee

$\begin{aligned} x^2 + x + 6 & = 62\\ x^2 + x + 6 - 62 & = 0\\ x^2 + x - 56 & = 0\\ x^2 + \left(8x - 7x\right) - 56 & = 0\\ \left(x^2 + 8x\right) - \left(7x + 56\right) & = 0\\ x\left(x + 8\right) - 7\left(x + 8\right) & = 0\\ \left(x + 8\right)\left(x - 7\right) & = 0 \end{aligned}$

$\begin{aligned} \left(x + 8\right) = 0 & \left(\text{or}\right) & \left(x - 7\right) = 0\\ x = -8 & \left(\text{or}\right) & x = 7 \end{aligned}$

So, the positive value of $x$ satisfying the given equation = $\color{#69047E}{\boxed{7}}$ .

First subtract $6$ from both side. That gives us $x^2+x=56$ . And $x^2+x=56$ . So $7^2+7=56$