Question

-x^2+16x+64=0 need to know using quadratic formula to find the zeroes

Answer 1

Answer:

$x= 8\left(1- \sqrt 2 \right)\\\\\\x= 8\left(1+ \sqrt 2 \right)$

Step by step explanation:

$-x^2 +16x +64 =0\\\\\text{Use the quadratic equation formula,}~~ x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{In this case,}~ a=-1, b=16, c=64\\\\\\x=\dfrac{-16\pm\sqrt{16^2 -4 \cdot (-1)(64)}}{2(-1)}\\\\\\~~~=\dfrac{-16\pm\sqrt{512}}{-2}\\\\\\~~~=\dfrac{-16 \pm \sqrt{256 \times 2}}{-2}\\\\\\~~~=\dfrac{-16 \pm 16\sqrt2}{-2}\\\\\\~~~=8 \pm(-8)\sqrt 2\\\\\text{Hence,}\\\\x=8-8\sqrt 2 = 8\left(1- \sqrt 2 \right)\\\\x= 8+8\sqrt 2 = 8\left(1+ \sqrt 2 \right)$