8x+10x+2x^2=440

Solution for 8x+10x+2x^2=440 equation:

8x+10x+2x^2=440

We move all terms to the left:

8x+10x+2x^2-(440)=0

We add all the numbers together, and all the variables

2x^2+18x-440=0

a = 2; b = 18; c = -440;
Δ = b2-4ac
Δ = 182-4·2·(-440)
Δ = 3844
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3844}=62$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-62}{2*2}=\frac{-80}{4}
=-20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+62}{2*2}=\frac{44}{4}
=11
$