9x^2+6x+16=14x^2+19x+6

Solution for 9x^2+6x+16=14x^2+19x+6 equation:

9x^2+6x+16=14x^2+19x+6

We move all terms to the left:

9x^2+6x+16-(14x^2+19x+6)=0

We get rid of parentheses

9x^2-14x^2+6x-19x-6+16=0

We add all the numbers together, and all the variables

-5x^2-13x+10=0

a = -5; b = -13; c = +10;
Δ = b2-4ac
Δ = -132-4·(-5)·10
Δ = 369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3\sqrt{41}}{2*-5}=\frac{13-3\sqrt{41}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3\sqrt{41}}{2*-5}=\frac{13+3\sqrt{41}}{-10}
$