3x^2+18x=19

Solution for 3x^2+18x=19 equation:

3x^2+18x=19

We move all terms to the left:

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

a = 3; b = 18; c = -19;
Δ = b2-4ac
Δ = 182-4·3·(-19)
Δ = 552
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:

$\sqrt{\Delta}=\sqrt{552}=\sqrt{4*138}=\sqrt{4}*\sqrt{138}=2\sqrt{138}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{138}}{2*3}=\frac{-18-2\sqrt{138}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{138}}{2*3}=\frac{-18+2\sqrt{138}}{6}
$