Question

$(2x - 4)(3x + 1)$

Answer 1

$(2x-4)(3x+1) = 6x^{2} +2x-12x -4 = 6x^{2} -10x -4$

a=6

b=-10

c=-4

Δ=$(-10)^{2} -4.6.-4$

Δ=$100+96$

Δ=$196$

$x1=\frac{+10+\sqrt{196}}{2.6} =\frac{10+14}{12} =\frac{24}{12} =2$

$x2=\frac{+10-\sqrt{196}}{2.6} =\frac{10-14}{12} =-\frac{4}{12} =-\frac{1}{3}$