Answer:
[tex](x-4)=0[/tex] or [tex](x+2)=0[/tex]
Step-by-step explanation:
we have
[tex](x + 1)(x - 3) = 5[/tex]
Multiply
[tex]x^{2} -3x+x-3=5[/tex]
[tex]x^{2} -2x-3-5=0[/tex]
[tex]x^{2} -2x-8=0[/tex]
we know that
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]x^{2} -2x-8=0[/tex]
so
[tex]a=1\\b=-2\\c=-8[/tex]
substitute in the formula
[tex]x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-8)}} {2(1)}[/tex]
[tex]x=\frac{2(+/-)\sqrt{36}} {2}[/tex]
[tex]x=\frac{2(+/-)6} {2}[/tex]
[tex]x=\frac{2+6} {2}=4[/tex]
[tex]x=\frac{2-6} {2}=-2[/tex]
therefore
[tex](x-4)(x+2)=0[/tex]
so
[tex](x-4)=0[/tex] or [tex](x+2)=0[/tex]
