440-2x=120-10x+.25x^2

Solution for 440-2x=120-10x+.25x^2 equation:

440-2x=120-10x+.25x^2

We move all terms to the left:

440-2x-(120-10x+.25x^2)=0

We get rid of parentheses

-.25x^2+10x-2x-120+440=0

We add all the numbers together, and all the variables

-0.25x^2+8x+320=0

a = -0.25; b = 8; c = +320;
Δ = b2-4ac
Δ = 82-4·(-0.25)·320
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{6}}{2*-0.25}=\frac{-8-8\sqrt{6}}{-0.5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{6}}{2*-0.25}=\frac{-8+8\sqrt{6}}{-0.5}
$