-6+3x-9x^2+16=0

Solution for -6+3x-9x^2+16=0 equation:

-6+3x-9x^2+16=0

We add all the numbers together, and all the variables

-9x^2+3x+10=0

a = -9; b = 3; c = +10;
Δ = b2-4ac
Δ = 32-4·(-9)·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{-(3)-3\sqrt{41}}{2*-9}=\frac{-3-3\sqrt{41}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{41}}{2*-9}=\frac{-3+3\sqrt{41}}{-18}
$