(x^2)+(16x)-9=0

Solution for (x^2)+(16x)-9=0 equation:

(x^2)+(16x)-9=0

a = 1; b = 16; c = -9;
Δ = b2-4ac
Δ = 162-4·1·(-9)
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{73}}{2*1}=\frac{-16-2\sqrt{73}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{73}}{2*1}=\frac{-16+2\sqrt{73}}{2}
$