$x^2+5x-7=0$
$x^2+2.\dfrac{5}{2}.x+\dfrac{25}{4}-\dfrac{53}{4}=0$
$x^2+2.\dfrac{5}{2}.x+(\dfrac{5}{2})^2-\dfrac{53}{4}=0$
$(x+\dfrac{5}{2})^2=\dfrac{53}{4}$
TH1: $x+\dfrac{5}{2}=\sqrt{\dfrac{53}{4}}$
$↔x=\sqrt{\dfrac{53}{4}}-\dfrac{5}{2}$
$↔x=\dfrac{\sqrt{53}}{2}-\dfrac{5}{2}$
$↔x=\dfrac{\sqrt{53}-5}{2}$
TH2: $x+\dfrac{5}{2}=-\sqrt{\dfrac{53}{4}}$
$↔x=-\sqrt{\dfrac{53}{4}}-\dfrac{5}{2}$
$↔x=-\dfrac{\sqrt{53}}{2}-\dfrac{5}{2}$
$↔x=\dfrac{-\sqrt{53}-5}{2}$
