2x^2+x+0.28=1

Solution for 2x^2+x+0.28=1 equation:

2x^2+x+0.28=1

We move all terms to the left:

2x^2+x+0.28-(1)=0

We add all the numbers together, and all the variables

2x^2+x-0.72=0

a = 2; b = 1; c = -0.72;
Δ = b2-4ac
Δ = 12-4·2·(-0.72)
Δ = 6.76
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{-(1)-\sqrt{6.76}}{2*2}=\frac{-1-\sqrt{6.76}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{6.76}}{2*2}=\frac{-1+\sqrt{6.76}}{4}
$