Question 987440
Original volume,  {{{3*5*8*cm^3}}}.


New volume using some added x to the first two dimensions, {{{(3+x)(5+x)*8=3*5*8+34}}}.
You essentially want to find x, the length addition to the first two dimensions.


STEPS TOWARD SOLUTION
{{{(3*5+5x+3x+x^2)8=3*5*8+34}}}
{{{3*5*8+64x+8x^2=3*5*8+34}}}
{{{8x^2+64x=34}}}
{{{highlight_green(4x^2+32x-17=0)}}}



Formula for general solution of a quadratic equation....
{{{Discriminant=32^2+4*4*17=1296=highlight_green(36^2)}}}


{{{x=(-32+36)/(2*4)}}}


{{{x=4/(2*4)}}}

{{{highlight(x=1/2)}}}------the length added each to 3 and the 5.