a. Write an equation that would be used to find three consecutive even integers such that the sum of the squares of the first and three times the third is ten more than six times the second.

Answer 1

let x-2, x and x+2 = the 3 consecutive even integers

then

(x-2)^2 + (3(x+2))^2 = 10+ 6x

x^2 -4x + 4 + 9x^2 + 36x +36 = 10+6x

10x^2 +26x +30 = 0

5x^2 +13x +15 =0

x=-13/10 + or - (1/10)sqr(169-300

but there's no integer or real solution so maybe you meant instead:

(x-2)^2 +3(x+2)^2 = 10+6x

x^2 -4x +4 + 3x^2 +12x +12 = 10+6x

4x^2 +2x +6 =0

2x^2 +x +3 =0

x=-1/4 + or - (1/2)sqr(1^2-24)

which has no real solution

there's a typo in your problem somewhere