8x^2-19x=-6

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