10x^2-19x+8=2

Solution for 10x^2-19x+8=2 equation:

10x^2-19x+8=2

We move all terms to the left:

10x^2-19x+8-(2)=0

We add all the numbers together, and all the variables

10x^2-19x+6=0

a = 10; b = -19; c = +6;
Δ = b2-4ac
Δ = -192-4·10·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*10}=\frac{8}{20}
=2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*10}=\frac{30}{20}
=1+1/2
$