$$(x+2)^2 = (x-5)^2+7$$

$$ \begin{aligned} (x+2)^2 &= (x-5)^2+7&& \text{simplify left and right hand side} \\[1 em]x^2+4x+4 &= x^2-10x+25+7&& \\[1 em]x^2+4x+4 &= x^2-10x+32&& \text{move all terms to the left hand side } \\[1 em]x^2+4x+4-x^2+10x-32 &= 0&& \text{simplify left side} \\[1 em]x^2+4x+4-x^2+10x-32 &= 0&& \\[1 em]14x-28 &= 0&& \text{ move the constants to the right } \\[1 em]14x &= 28&& \text{ divide both sides by $ 14 $ } \\[1 em]x &= \frac{28}{14}&& \\[1 em]x &= 2&& \\[1 em] \end{aligned} $$

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