9x^2-42x+K=0

Solution for 9x^2-42x+K=0 equation:

9x^2-42x+=0

We add all the numbers together, and all the variables

9x^2-42x=0

a = 9; b = -42; c = 0;
Δ = b2-4ac
Δ = -422-4·9·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-42}{2*9}=\frac{0}{18}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+42}{2*9}=\frac{84}{18}
=4+2/3
$