Question

dau 53 de puncte pentru raspunsul la aceaste probleme

Answer 1

3x^2+2x-5=0
x=\frac{-2+\sqrt{2^2-4\cdot \:3\left(-5\right)}}{2\cdot \:3}
=\frac{-2+\sqrt{2^2-\left(-5\right)\cdot \:3\cdot \:4}}{6}
\sqrt{2^2-4\cdot \:3\left(-5\right)}=\sqrt{64}
=\frac{-2+\sqrt{64}}{6}
\sqrt{64}=8
=\frac{-2+8}{6}
=1
x=\frac{-2-\sqrt{2^2-4\cdot \:3\left(-5\right)}}{2\cdot \:3}
\frac{-2-\sqrt{2^2-4\cdot \:3\left(-5\right)}}{2\cdot \:3}
=\frac{-2-\sqrt{2^2-\left(-5\right)\cdot \:3\cdot \:4}}{6}
\sqrt{2^2-4\cdot \:3\left(-5\right)}=\sqrt{64}
=\frac{-2-\sqrt{64}}{6}
\sqrt{64} = 8

=\frac{-2-8}{6}

=\frac{-10}{6}

=-\frac{10}{6}

=-\frac{5}{3}

x=1,  x=-\frac{5}{3}