3x^2+24x-11=19

Solution for 3x^2+24x-11=19 equation:

3x^2+24x-11=19

We move all terms to the left:

3x^2+24x-11-(19)=0

We add all the numbers together, and all the variables

3x^2+24x-30=0

a = 3; b = 24; c = -30;
Δ = b2-4ac
Δ = 242-4·3·(-30)
Δ = 936
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:

$\sqrt{\Delta}=\sqrt{936}=\sqrt{36*26}=\sqrt{36}*\sqrt{26}=6\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{26}}{2*3}=\frac{-24-6\sqrt{26}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{26}}{2*3}=\frac{-24+6\sqrt{26}}{6}
$