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I need a help!Please solve this question r2+200r-323 = 0

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Given:


r^2+200r-323=0
a=1\text{ ; b=200 ; c=-323}
r=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}
r=\frac{-200\pm\sqrt[]{200^2-4(1)(-323)}}{2(1)}
r=\frac{-200\pm\sqrt[]{400^{}+1292}}{2}
r=\frac{-200\pm\sqrt[]{400^{}+1292}}{2}
r=\frac{-200\pm\sqrt[]{1692}}{2}
r=\frac{-200\pm\sqrt[]{2^2*3^2*47}}{2}
r=\frac{-200\pm6\sqrt[]{47}}{2}
r=\frac{2(-100\pm3\sqrt[]{47})}{2}
r=-100\pm3\sqrt[]{47}
r=-100+3\sqrt[]{47}\text{ ,}-100-3\sqrt[]{47}

User Mihails Butorins
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