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\begin{aligned} (a + b)^2 &= c^2 + 4 \cdot \left( \frac{1}{2} ab \right) \\ a^2 + 2ab + b^2 &= c^2 + 2ab \\ a^2 + b^2 &= c^2 \end{aligned} \frac{1}{2} ab \right) \\ a^2 + 2ab + b^2 &= c^2 + \frac{1}{2} ab \right) \\ a^2 + 2ab + b^2 &= c^2 + 2ab \\ a^2 + b^2 &= c^2 \end{aligned} \frac{1}{2} ab \right) \\ a^2 + 2ab + b^2 &= c^2 + \frac{1}{2} ab \right) \\ a^2 + 2ab + b^2 &= c^2 + 2ab \\ a^2 + b^2 &= c^2 \end{aligned} \frac{1}{2} ab \right) \\ a^2 + 2ab + b^2 &= c^2 + \frac{1}{2} ab \right) \\ a^2 + 2ab + b^2 &= c^2 + 2ab \\ a^2 + b^2 &= c^2 \end{aligned} \frac{1}{2} ab \right) \\ a^2 + 2ab + b^2 &= c^2 + 2ab \\ a^2 + b^2 &= c^2 \end{aligned}
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MathJax via \(\TeX\) and JavaScript:
When \(a \ne 0\), two solutions to \(ax^2 + bx + c = 0\): \[x = {-b \pm \sqrt{b^2-4ac} \over 2a}\]
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