3a^2=-19+18a

Solution for 3a^2=-19+18a equation:

3a^2=-19+18a

We move all terms to the left:

3a^2-(-19+18a)=0

We add all the numbers together, and all the variables

3a^2-(18a-19)=0

We get rid of parentheses

3a^2-18a+19=0

a = 3; b = -18; c = +19;
Δ = b2-4ac
Δ = -182-4·3·19
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{6}}{2*3}=\frac{18-4\sqrt{6}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{6}}{2*3}=\frac{18+4\sqrt{6}}{6}
$