1x+2x^2=28

Solution for 1x+2x^2=28 equation:

1x+2x^2=28

We move all terms to the left:

1x+2x^2-(28)=0

We add all the numbers together, and all the variables

2x^2+x-28=0

a = 2; b = 1; c = -28;
Δ = b2-4ac
Δ = 12-4·2·(-28)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*2}=\frac{14}{4}
=3+1/2
$