Answer:
[tex]26[/tex] and [tex]28[/tex].
Step-by-step explanation:
Let [tex]x[/tex] denote the smaller one of the two even integers ([tex]x > 0[/tex] since both integers are positive.) The larger one of the two consecutive even integer would be [tex](x + 2)[/tex].
The square of the smaller integer would be [tex]x^{2}[/tex].
Five times the larger integer would be [tex]5\, (x + 2)[/tex].
Subtract five times the larger integer from the square of the smaller integer to get [tex](x^{2} - 5\, (x + 2))[/tex].
The value of this expression should be equal to [tex]536[/tex]. In other words:
[tex]x^{2} - 5\, (x + 2) = 536[/tex].
Rewrite and simplify this quadratic equation:
[tex]x^{2} + (-5)\, x + (- 546) = 0[/tex].
Apply the quadratic formula to find possible values of [tex]x[/tex]:
[tex]\begin{aligned}x_{1} &= \frac{-b + \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(-5) + \sqrt{(-5)^{2} - 4 \times 1 \times (-546)}}{2}\\ &= \frac{5 + \sqrt{2209}}{2} \\ &=\frac{5 + 47}{2} \\ &= 26\end{aligned}[/tex].
[tex]\begin{aligned}x_{2} &= \frac{-b - \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{5 - \sqrt{2209}}{2} \\ &=\frac{5 - 47}{2} \\ &= -21\end{aligned}[/tex].
Since [tex]x > 0[/tex] (both numbers are supposed to be positive), [tex]x = 26[/tex] would be the only valid solution.
Therefore, the two integers would be [tex]x = 26[/tex] and [tex]x + 2 = 28[/tex].
