16x2+39x+18=0

Solution for 16x2+39x+18=0 equation:

16x^2+39x+18=0

a = 16; b = 39; c = +18;
Δ = b2-4ac
Δ = 392-4·16·18
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-3\sqrt{41}}{2*16}=\frac{-39-3\sqrt{41}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+3\sqrt{41}}{2*16}=\frac{-39+3\sqrt{41}}{32}
$