[tex]\displaystyle\bf 4x^2+3=-7x\\\\4x^2+7x+3=0 \\D=49-48=1 \\\\x_{1;2}=\frac{-7\pm1}{8} => x_1=-1\quad ; \quad x_2=-\frac{3}{4} \\\\\\Answer:\boxed{x_1=-1\quad ; \quad x_2=-\frac{3}{4} }[/tex]
