16x^2-42x+16=0

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

16x^2-42x+16=0

a = 16; b = -42; c = +16;
Δ = b2-4ac
Δ = -422-4·16·16
Δ = 740
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:

$\sqrt{\Delta}=\sqrt{740}=\sqrt{4*185}=\sqrt{4}*\sqrt{185}=2\sqrt{185}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{185}}{2*16}=\frac{42-2\sqrt{185}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{185}}{2*16}=\frac{42+2\sqrt{185}}{32}
$