6x^2+19x-32=4

Solution for 6x^2+19x-32=4 equation:

6x^2+19x-32=4

We move all terms to the left:

6x^2+19x-32-(4)=0

We add all the numbers together, and all the variables

6x^2+19x-36=0

a = 6; b = 19; c = -36;
Δ = b2-4ac
Δ = 192-4·6·(-36)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-35}{2*6}=\frac{-54}{12}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+35}{2*6}=\frac{16}{12}
=1+1/3
$