Question

Ecuații de gradul al doilea
3x²+x-4=0​

Answer 1

Salut!

$3x^{2}+x-4=0$

Δ=$b^{2}-4ac$

Δ=$1^{2}+4\times3\times4$

Δ=$1+48$

Δ=$49$

$x_{1}=\frac{-1+\sqrt49}{2\times3}\\x_{2}=\frac{-1-\sqrt49}{2\times3}$

$x_1=\frac{6}{6}=1;\\x_2=\frac{-8}{6}=-\frac{4}{3}$

Bafta, si sper ca ai inteles!