16q^2+218q+574=0

Solution for 16q^2+218q+574=0 equation:

16q^2+218q+574=0

a = 16; b = 218; c = +574;
Δ = b2-4ac
Δ = 2182-4·16·574
Δ = 10788
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{10788}=\sqrt{4*2697}=\sqrt{4}*\sqrt{2697}=2\sqrt{2697}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(218)-2\sqrt{2697}}{2*16}=\frac{-218-2\sqrt{2697}}{32}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(218)+2\sqrt{2697}}{2*16}=\frac{-218+2\sqrt{2697}}{32}
$