Ettiene solved a quadratic equation. His work is shown below. In which step did Ettiene make an error? \begin{aligned} \dfrac{3}{4}(x-6)^2+1&=28 \dfrac{3}{4}(x-6)^2&=27&\te xt{Step }1 (x-6)^2&=20\dfrac14&\text{S tep }2 x-6&=\pm\sqrt{20\dfrac14}&\ text{Step }3 x&=\pm\sqrt{20\dfrac14}+6&\ text{Step }4 \end{aligned} 4 3 (x−6) 2 +1 4 3 (x−6) 2 (x−6) 2 x−6 x =28 =27 =20 4 1 =± 20 4 1 =± 20 4 1 +6 Step 1 Step 2 Step 3 Step 4