## Calculus with Applications (10th Edition)

Set the $RHS$ to 0 by subtracting $14m$ and adding 49 to both sides: $m^{2}=14m-49 \qquad.../-14m+49$ $m^{2}-14m+49=0$ In R-2, see "Perfect square" in Special Factorizations. $(A+B)^{2}=A^{2}+2AB+B^{2}$ $(A-B)^{2}=A^{2}-2AB+B^{2}$ $m^{2}$ is the square of m, 49 is the square of 7, and 14m = 2(m)(7). Our equation becomes $(m-7)^{2}=0$ which can only be if $m-7=0,\qquad$ so $m=7$