LilasU656985
LilasU656985
30-10-2022
Mathematics
contestada
How to solve 5x^2-45=0
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KelaiahX112423
KelaiahX112423
30-10-2022
[tex]\begin{gathered} 5x^2-45=0 \\ a=5 \\ b=0 \\ c=-45 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-0\pm\sqrt[]{0^2-4(5)(-45)}}{2(5)} \\ x=\frac{\pm\sqrt[]{900}}{10} \\ x=\frac{\pm30}{10} \\ x=\pm3 \end{gathered}[/tex]
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