4x^2-18x-16=0

Solution for 4x^2-18x-16=0 equation:

4x^2-18x-16=0

a = 4; b = -18; c = -16;
Δ = b2-4ac
Δ = -182-4·4·(-16)
Δ = 580
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:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{145}}{2*4}=\frac{18-2\sqrt{145}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{145}}{2*4}=\frac{18+2\sqrt{145}}{8}
$