Algebra

Sorry, I've just noticed an error with my answer. The starting equation should simplify to 2x^2 - 4x - 68. You can use the same process to find out that the roots are 1 + sqrt(35) and 1 - sqrt(35)

Full process:

$2x^2-4x-68=0$

$x^2-2x-34=0$

$x^2-2x+1=35$

$(x-1)^2 = 35$

$x-1 = \pm \sqrt{35}$

$x = 1 \pm \sqrt{35}$

We can first simplify this equation by combining like terms

$x^2+x^2-11x+2x+5x-42-10-16 = 0$

$2x^2+4x-68 = 0$

$x^2+4x-34=0$

Now we can use completing the square to solve this polynomial equation.

$x^2+4x -34 + 38 = 0 + 38$

$x^2+4x+4 = 38$

$(x+2)^2 = 38$

$x + 2 = \pm \sqrt{38}$

$x = -2 \pm \sqrt{38}$

So the 2 solutions are -2 + sqrt(38) and -2 - sqrt(38)