Question

solutions of equation z^2-12z+36=0

Answer 1

Using the quadratic equations roots formula,\[(-b \pm \sqrt{b^2-4ac})/2a\]for\[ax^2+bx+c\]
\[(-(-12) \pm \sqrt{(-12)^2-4(1)(36)})/2(1)\]\[(12 \pm \sqrt{(144-144})/2\]\[(12 \pm \sqrt{(0})/2\]\[(12 \pm 0)/2\]\[12/2\]\[z=6\]

Answer 2

z^2-12z+36=0
=>z^2-6z-6z+36=0
=>z(z-6)-6(z-6)=0
=>(z-6)(z-6)=0
z=6,6