Question

Solve
${(4x+3)}^2=16$

Answer 1

Given,

${(4x+3)}^2=16$

$\Rightarrow (4x+3{)}^2-4^2=0$

$\Rightarrow (4x+3-4)(4x+3+4)=0$

$\Rightarrow (4x-1)(4x+7)=0$

$\Rightarrow x=\frac{1}{4},-\frac{7}{4}$.