Answer:
[tex]x=\frac{-1(+)\sqrt{10}} {3}[/tex]
[tex]x=\frac{-1(-)\sqrt{10}} {3}[/tex]
[tex]x=1[/tex]
Step-by-step explanation:
we have
[tex](3x^{2}+2x-3)(x-1)[/tex]
Solve the quadratic equation
[tex](3x^{2}+2x-3)[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]3x^{2}+2x-3=0[/tex]
so
[tex]a=3\\b=2\\c=-3[/tex]
substitute in the formula
[tex]x=\frac{-2(+/-)\sqrt{2^{2}-4(3)(-3)}} {2(3)}[/tex]
[tex]x=\frac{-2(+/-)\sqrt{40}} {6}[/tex]
[tex]x=\frac{-2(+/-)2\sqrt{10}} {6}[/tex]
[tex]x=\frac{-2(+)2\sqrt{10}} {6}=\frac{-1(+)\sqrt{10}} {3}[/tex]
[tex]x=\frac{-2(-)2\sqrt{10}} {6}=\frac{-1(-)\sqrt{10}} {3}[/tex]
therefore
The solutions of the equation
[tex](3x^{2}+2x-3)(x-1)[/tex]
are
[tex]x=\frac{-1(+)\sqrt{10}} {3}[/tex]
[tex]x=\frac{-1(-)\sqrt{10}} {3}[/tex]
[tex]x=1[/tex]
