Question

\(4x^2+3x-3=0\)

Answer 1

\(4x^2+3x-3=0\\4(x^2 +3/4 x+9/16)-21/4=0\\4(x+3/4)^2 =21/4\\(x+3/4)^2 =21/16\\x+3/4=±\dfrac{\sqrt{21}}{4}\\x=\dfrac{±\sqrt{21}-3}{4}\)

Answer 2

\(4x^2+3x-3=0\)

\(a=4;b=3;c=-3\)

\(x=\sqrt{\dfrac{-3\pm\sqrt{3^2-4\times4\times\left(-3\right)}}{2\times4}}\)

\(x=\dfrac{-3\pm\sqrt{9+48}}{8}\)

\(x=\dfrac{-3\pm\sqrt{57}}{8}\)

\(x=\dfrac{-3+\sqrt{57}}{8}\)

\(x=\dfrac{-3-\sqrt{57}}{8}\)

\(x_1=\dfrac{-3-\sqrt{57}}{8};x_2=\dfrac{-3+\sqrt{57}}{8}\)

pt có 2 nghiệm.