If ɑ and β are roots of 2x² - 5x + 7, then
2x² - 5x + 7 = 2 (x - ɑ) (x - β)
Expanding the right side and matching up coefficients, it follows that
2ɑβ = 7 ⇒ ɑβ = 7/2
-2 (ɑ + β) = -5 ⇒ ɑ + β = 5/2
Now consider a polynomial whose roots are 3ɑ + 4β and 4ɑ + 3β, which can be written with factorization
(x - (3ɑ + 4β)) (x - (4ɑ + 3β))
and expanded to get
x² - (7ɑ + 7β) x + (3ɑ + 4β) (4ɑ + 3β)
Expand/factor the coefficients to rewrite them in terms of the known relations above.
x² - 7 (ɑ + β) x + (12ɑ² + 25ɑβ + 12β²)
x² - 7 (ɑ + β) x + 12 (ɑ² + β²) + 25ɑβ
Recall the square of a binomial,
(x + y)² = x² + 2xy + y²
so we rewrite the latter quadratic as
x² - 7 (ɑ + β) x + 12 ((ɑ + β)² - 2ɑβ) + 25ɑβ
x² - 7 (ɑ + β) x + 12 (ɑ + β)² + ɑβ
Then substituting the expressions whose values we know, we get
x² - 7 (5/2) x + 12 (5/2)² + 7/2
= x² - 35/2 x + 157/2
which we can multiply by 2 to get integer coefficients like in the given quadratic,
= 2x² - 35x + 157