Question

Solve : $16x^2+8x+1$

Answer 1

$16x^2+8x+1=0$

$\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$b=8$ $a=16$ $c=1$

$\Rightarrow x=\frac{-8\pm \sqrt{64-4\left(16\right)\left(1\right)}}{2\left(16\right)}$

$=\frac{-8\pm \sqrt{0}}{2\left(16\right)}$

$\Rightarrow \frac{-8}{32}\Rightarrow \frac{-1}{4}$

$\therefore x=\frac{-1}{4}$