0=160-40x^2+168x

Solution for 0=160-40x^2+168x equation:

0=160-40x^2+168x

We move all terms to the left:

0-(160-40x^2+168x)=0

We add all the numbers together, and all the variables

-(160-40x^2+168x)=0

We get rid of parentheses

40x^2-168x-160=0

a = 40; b = -168; c = -160;
Δ = b2-4ac
Δ = -1682-4·40·(-160)
Δ = 53824
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{53824}=232$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-168)-232}{2*40}=\frac{-64}{80}
=-4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-168)+232}{2*40}=\frac{400}{80}
=5
$