r^2+18r-43=0

Solution for r^2+18r-43=0 equation:

r^2+18r-43=0

a = 1; b = 18; c = -43;
Δ = b2-4ac
Δ = 182-4·1·(-43)
Δ = 496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-4\sqrt{31}}{2*1}=\frac{-18-4\sqrt{31}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+4\sqrt{31}}{2*1}=\frac{-18+4\sqrt{31}}{2}
$