27x^2-42x+16=0

Solution for 27x^2-42x+16=0 equation:

27x^2-42x+16=0

a = 27; b = -42; c = +16;
Δ = b2-4ac
Δ = -422-4·27·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6}{2*27}=\frac{36}{54}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6}{2*27}=\frac{48}{54}
=8/9
$