Question

ASAP WILL MARK BRAINLYEST solve this equation by completing the square:

Answer 1

$(*)\qquad(a+b)^2=a^2+2ab+b^2\\\\x^2+8x-13=0\ \ \ \ |+13\\\\x^2+2\cdot x\cdot4=13\ \ \ \ |+4^2\\\\\underbrace{x^2+2\cdot x\cdot4+4^2}_{(*)}=13+4^2\\\\(x+4)^2=29\to x+4=\pm\sqrt{29}\ \ \ \ |-4\\\\\boxed{x=-4-\sqrt{29}\ \vee\ x=-4+\sqrt{29}}$