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Solve for r. r^2 +3r=7
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-b +- root(b^2 -4(a)(c)) all over 2(a) with a=1 b=3 c=-7
\[r^2+3r-7=0\] use \[r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \[a=1,b=3,c=-7\]
the answer was -3+sqrt37/2, -3-sqrt37/2 I'm confused
notice in satellite's post that when you substitute into the quadratic formula, you get two possible answers. for the portion of the formula b^2-4ac = (3^2) -4*1*-7=9+28=37 so we get \[r= (-3\pm \sqrt{37})/ 2\]
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