0=5x2-42x+16

Solution for 0=5x2-42x+16 equation:

0=5x^2-42x+16

We move all terms to the left:

0-(5x^2-42x+16)=0

We add all the numbers together, and all the variables

-(5x^2-42x+16)=0

We get rid of parentheses

-5x^2+42x-16=0

a = -5; b = 42; c = -16;
Δ = b2-4ac
Δ = 422-4·(-5)·(-16)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-38}{2*-5}=\frac{-80}{-10}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+38}{2*-5}=\frac{-4}{-10}
=2/5
$