x^2+16x-43=0

Solution for x^2+16x-43=0 equation:

x^2+16x-43=0

a = 1; b = 16; c = -43;
Δ = b2-4ac
Δ = 162-4·1·(-43)
Δ = 428
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:

$\sqrt{\Delta}=\sqrt{428}=\sqrt{4*107}=\sqrt{4}*\sqrt{107}=2\sqrt{107}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{107}}{2*1}=\frac{-16-2\sqrt{107}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{107}}{2*1}=\frac{-16+2\sqrt{107}}{2}
$