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how to solve r^2+3r=5?

r=

4 Answers

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  • 7 years ago
    Favorite Answer

    R= 1.1926, -4.1926

    Try using MathWay.com

    Hope this helps!

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  • DWRead
    Lv 7
    7 years ago

    r² + 3r = 5

    set to 0

    r² + 3r - 5 = 0

    quadratic formula

    r = [-3 ± √(3² + 4·1·5)]/(2·1)

     = [-3 ± √29]/2

     = -1.5 ± 0.5√29

     ≅ -4.193, 1.193

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  • 7 years ago

    r^2+3r-5=0

    r=[-3+-sqrt(9+20)]/2

    r1=(-3+5.385)/2=1.1925

    r2=(-3-5.385)/2=-4.1925

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  • JOS J
    Lv 7
    7 years ago

    {r = 1/2 (-3 - Sqrt[29])}, {r = 1/2 (-3 + Sqrt[29])}

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