r^2+18r-46=0

Solution for r^2+18r-46=0 equation:

r^2+18r-46=0

a = 1; b = 18; c = -46;
Δ = b2-4ac
Δ = 182-4·1·(-46)
Δ = 508
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:

$\sqrt{\Delta}=\sqrt{508}=\sqrt{4*127}=\sqrt{4}*\sqrt{127}=2\sqrt{127}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{127}}{2*1}=\frac{-18-2\sqrt{127}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{127}}{2*1}=\frac{-18+2\sqrt{127}}{2}
$