3a2+8a-19=-3

Solution for 3a2+8a-19=-3 equation:

3a^2+8a-19=-3

We move all terms to the left:

3a^2+8a-19-(-3)=0

We add all the numbers together, and all the variables

3a^2+8a-16=0

a = 3; b = 8; c = -16;
Δ = b2-4ac
Δ = 82-4·3·(-16)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*3}=\frac{-24}{6}
=-4
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*3}=\frac{8}{6}
=1+1/3
$