Question

$4 {x}^{2} - 12x + 8 = 0$
va rooog
E urgent

Answer 1

$4 {x}^{2} - 12x + 8 = 0$

$a = 4$

$b = - 12$

$c = 8$

$\Delta = {b}^{2} - 4ac$

$\Delta = {( - 12)}^{2} - 4 \times 4 \times 8$

$\Delta = 144 - 128$

$\Delta = 16 > 0 = > \exists \: x_{1} \: \neq \: x_{2}$

$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$

$x_{1,2}=\frac{-( - 12)\pm\sqrt{16}}{2 \times 4}$

$x_{1,2}=\frac{12\pm4}{8}$

$x_{1}=\frac{12 + 4}{8}$

$x_{1}=\frac{16}{8}$

$x_{1}=2$

$x_{2}=\frac{12 - 4}{8}$

$x_{2}=\frac{8}{8}$

$x_{2}=1$