Answer :

mariamikayla
[tex]4x^2+28x-32=0 \ \ \ |:4 \\\\ x^2+7x-8=0 \\\\ x^2+8x-x-8=0 \\\\ x(x+8)-(x+8)=0 \\\\ (x+8)(x-1)=0 \\\\ \\x+8=0 \\\\ \boxed{x=-8} \\\\\\ x-1=0 \\\\ \boxed{x=1} \\\\\\ \boxed{\boxed{S=\{-8;1 \}}}[/tex]
apologiabiology
we know that if xy=0 then x or/and y=0


4x^2+28x-32=0
factor out the 4
4(x^2+7-8)=0

factor out the x^2+7-8
we must find out what two numbers multiply to get -8 and add to get 7
the numbers are 8 and -1

so 4[(x-1)(x+8)]=0

we know that if exg 4y=0 then y=0
so
therefor
(x-1)(x+8)=0
set each to zero
x-1=0
x=1
x+8=0
x=-8

x could be 1 or -8

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