4x^2+19x+12=-5

Solution for 4x^2+19x+12=-5 equation:

4x^2+19x+12=-5

We move all terms to the left:

4x^2+19x+12-(-5)=0

We add all the numbers together, and all the variables

4x^2+19x+17=0

a = 4; b = 19; c = +17;
Δ = b2-4ac
Δ = 192-4·4·17
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{89}}{2*4}=\frac{-19-\sqrt{89}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{89}}{2*4}=\frac{-19+\sqrt{89}}{8}
$