$(x-1)^2+5=(2x+3)^2(1)$
$ĐKXĐ:x∈R$
$pt(1)⇔x^2-2x+1+5=4x^2+12x+9$
$⇔4x^2-x^2+12x+2x+9-1-5=0$
$⇔3x^2+14x+3=0$
$⇔3(x^2+\dfrac{14}{3}x+1)=0$
$⇔(x^2+\dfrac{14}{3}x+1)=0$
$⇔x^2+2.\dfrac{14}{6}x+\dfrac{196}{36}=\dfrac{160}{36}$
$⇔(x+\dfrac{14}{6})^2=\dfrac{40}{9}$
$⇔(x+\dfrac{14}{6})^2-\dfrac{40}{9}$=0$
$⇔(x+\dfrac{14}{6}-\dfrac{\sqrt[]{40}}{3}).(x+\dfrac{14}{6}+\dfrac{\sqrt[]{40}}{3})=0$
$⇔$\(\left[ \begin{array}{l}x+\dfrac{7}{3}=\dfrac{\sqrt[]{40}}{3}\\x+\dfrac{7}{3}=\dfrac{-\sqrt[]{40}}{3}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{\sqrt[]{40}-7}{3}\\x=\dfrac{-\sqrt[]{40}-7}{3}\end{array} \right.\)
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