2t2t^{2}t2+2 $\sqrt{\3}$-3
asked by guest on Nov 15, 2024 at 12:54 am
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MathBot Answer:
2t2+23−3=2t2−3+232 {t}^{2} + 2 \sqrt{3} - 3 = 2 t^{2} - 3 + 2 \sqrt{3}2t2+23−3=2t2−3+23