# Consider the following function r(x)=-(x+5)^2+2

Consider the following function
$r\left(x\right)=-\left(x+5{\right)}^{2}+2$
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Cremolinoer
Given $r\left(x\right)=-\left(x+5{\right)}^{2}+2$
so
$0=-\left(x+5{\right)}^{2}+2\phantom{\rule{0ex}{0ex}}-\left(x+5{\right)}^{2}+2=0\phantom{\rule{0ex}{0ex}}-\left({x}^{2}+2x×5+{5}^{2}\right)+2=0\phantom{\rule{0ex}{0ex}}-{x}^{2}-10x-25+2=0\phantom{\rule{0ex}{0ex}}-{x}^{2}-10-23=0\phantom{\rule{0ex}{0ex}}{x}^{2}+10x+23=0$
Now we can solve this equation using quadratic formula
$x=\frac{-10±\sqrt{{10}^{2}-4×23}}{2×1}\phantom{\rule{0ex}{0ex}}x=-10±\frac{\sqrt{100-92}}{2}\phantom{\rule{0ex}{0ex}}x=-10±\frac{\sqrt{8}}{2}\phantom{\rule{0ex}{0ex}}x=-10±\frac{\sqrt{4×2}}{2}\phantom{\rule{0ex}{0ex}}x=2\left(\frac{-5±\sqrt{2}}{2}\right)\phantom{\rule{0ex}{0ex}}x=-5±\sqrt{2}$