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completing the square
v^2 + 4v - 22= 10


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[tex]v^2+4v-22=10 \\ v^2+4v+4-4-22=10 \\ (v+2)^2-26=10 \\ (v+2)^2=36 \\ \sqrt{(v+2)^2}=\sqrt{36} \\ |v+2|=6 \\ v+2=6 \ \lor \ v+2=-6 \\ v=4 \ \lor \ v=-8 \\ \boxed{v=-8 \hbox{ or } v=4}[/tex]
v² + 4v - 22 = 10
            - 10  - 10
v² + 4v - 32 = 0
v = -4 +/- √(4² - 4(1)(-32))
                     2(1)
v = -4 +/- √(16 + 128)
                    2
v = -4 +/- √(144)
                2
v = -4 +/- 12
             2
v = -2 +/- 6
v = -2 + 6    ∨    v = -2 - 6
v = 4            ∨    v = -8