## Algebra 1

$w=13,1$
$w^2-14w+13=0$ or, $w^2-14w=-13$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=-14$ Therefore, $b^2=4ac$ $\implies$ $c=\dfrac{b^2}{4a}$ Thus, $c=\dfrac{b^2}{4a}=\dfrac{(-14)^2}{4}=49$ To complete the square, add $4$ on both sides. $w^2-14w+49=-13+49$ $\implies (w-7)^2=36$ $\implies w-7=6$ and $\implies w-7=-6$ or, $w=13,1$