3x^2+10x-19=11-3x

Solution for 3x^2+10x-19=11-3x equation:

3x^2+10x-19=11-3x

We move all terms to the left:

3x^2+10x-19-(11-3x)=0

We add all the numbers together, and all the variables

3x^2+10x-(-3x+11)-19=0

We get rid of parentheses

3x^2+10x+3x-11-19=0

We add all the numbers together, and all the variables

3x^2+13x-30=0

a = 3; b = 13; c = -30;
Δ = b2-4ac
Δ = 132-4·3·(-30)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-23}{2*3}=\frac{-36}{6}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+23}{2*3}=\frac{10}{6}
=1+2/3
$