Answer:
12x^2+2x+7
Step-by-step explanation:
Distribute the Negative Sign:
=4x^2+6+7+−1(−8x^2−2x+6)
=4x^2+6+7+−1(−8x^2)+−1(−2x)+(−1)(6)
=4x^2+6+7+8x^2+2x+−6
Combine Like Terms:
=4x^2+6+7+8x^2+2x+−6
=(4x^2+8x^2)+(2x)+(6+7+−6)
=12x^2+2x+7

so we have a 33, namely two real solutions for that quadratic.
usually that number goes into a √, if you have covered the quadratic formula, you'd see it there, namely that'd be equivalent to √(33), now 33 is a prime number, and √(33) is yields an irrational value, specifically because a prime number is indivisible other than by itself or 1.
so 33 can only afford us two real irrational roots.
3x + 1 + 5x = 7 + 15 + 7x
8x + 1 = 22 + 7x
x = 21