4x^2+19x+3=8

Solution for 4x^2+19x+3=8 equation:

4x^2+19x+3=8

We move all terms to the left:

4x^2+19x+3-(8)=0

We add all the numbers together, and all the variables

4x^2+19x-5=0

a = 4; b = 19; c = -5;
Δ = b2-4ac
Δ = 192-4·4·(-5)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*4}=\frac{-40}{8}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*4}=\frac{2}{8}
=1/4
$