0=x^2-16x+42

Solution for 0=x^2-16x+42 equation:

0=x^2-16x+42

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-x^2+16x-42=0

We add all the numbers together, and all the variables

-1x^2+16x-42=0

a = -1; b = 16; c = -42;
Δ = b2-4ac
Δ = 162-4·(-1)·(-42)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{22}}{2*-1}=\frac{-16-2\sqrt{22}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{22}}{2*-1}=\frac{-16+2\sqrt{22}}{-2}
$