Question 858033
Let:

{{{x^2}}}, {{{(x+1)^2}}} and {{{(x+2)^2}}} be the three consecutive perfect square.

{{{x^2+(x+1)^2+(x+2)^2=2030}}}
{{{x^2+(x^2+2x+1)+(x^2+4x+4)=2030}}}
{{{3x^2+6x+5=2030}}}
{{{3x^2+6x=2025}}}
{{{x^2+2x=675}}}
{{{x^2+2x-675=0}}}
{{{(x+27)(x-25)=0}}}
{{{x+27=0}}}
{{{x=-27}}}
{{{x-25=0}}}
{{{x=25}}}


So the perfect squares are...
25^2    = 625
25+1^2  = 676
25+2^2  = 729

Which all in all totals to 2030!



HOPE THAT HELPS!

Follow me on insta and twitter @seutip that'd be a great help!


<3