Question

Ac method for x^2+20x-8=0

Answer 1

Step-by-step explanation:

By using quadratic formula:

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

`\quad a=1,\:b=20,\:c=-8`

`x_(1,\:2)=(-20\pm √(20^2-4\cdot \:1\left(-8\right)))/(2\cdot \:1)`

`x=(-20+√(20^2-4\cdot \:1\left(-8\right)))/(2\cdot \:1):\quad 2\left(3√(3)-5\right)`

`x=(-20-√(20^2-4\cdot \:1\left(-8\right)))/(2\cdot \:1):\quad -2\left(5+3√(3)\right)`

So the solutions are:

`\boxed{x=2\left(3√(3)-5\right),\:x=-2\left(5+3√(3)\right)}`