Let's solve in $ℝ$: $x^2+\frac{8}{7}+\frac{15}{49}=0$

ⓘ To add fractions, put them over the same denominator, then add the numerators. We choose $49$, the least common multiple of $7$ and $49$. Multiplying the numerator and the denominator by the same number does not change the fraction: $\frac{8}{7}+\frac{15}{49}$ = $\frac{8·7}{7·7}+\frac{15}{49}$ = $\frac{8·7+15}{49}$

$x^2+\frac{8·7+15}{49}=0$

ⓘ Multiplications are done before additions and subtractions: $8·7$ = $56$

$x^2+\frac{56+15}{49}=0$

ⓘ The numerator and the denominator are computed before the division: $56+15$ = $71$

$x^2+\frac{71}{49}=0$

$x^2=-\frac{71}{49}$

In ℝ, a square is never negative: no real number has $-\frac{71}{49}$ as its square

▶Solution: $x\in \varnothing$