Question

(5x²-4x+7)-(3x²+7x-1)=1

Answer 1

$(5x^2-4x+7)-(3x^2+7x-1)=1\\\\ 5x^2-4x+7-3x^2-7x+1-1=0\\\\ 5x^2-3x^2-4x-7x+7=0\\\\ 2x^2-11x+7=0$

Encontrando as raízes utilizando a fórmula de Bháskara:
$a = 2\ \ \ \ b=-11\ \ \ \ c=7\\\\\\ x = \dfrac{-b\pm\sqrt{b^2-4.a.c}}{2.a}\\\\ x = \dfrac{11\pm\sqrt{121-56}}{4}\\\\ x = \dfrac{11\pm\sqrt{65}}{4}\\\\ \boxed{x'=\dfrac{11+\sqrt{65}}{4}}\\\\\\ \boxed{x''=\dfrac{11-\sqrt{65}}{4}}$

Bons estudos!