8x2+19x=-6

Solution for 8x2+19x=-6 equation:

8x^2+19x=-6

We move all terms to the left:

8x^2+19x-(-6)=0

We add all the numbers together, and all the variables

8x^2+19x+6=0

a = 8; b = 19; c = +6;
Δ = b2-4ac
Δ = 192-4·8·6
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*8}=\frac{-32}{16}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*8}=\frac{-6}{16}
=-3/8
$