r^2+10r-19=0

Solution for r^2+10r-19=0 equation:

r^2+10r-19=0

a = 1; b = 10; c = -19;
Δ = b2-4ac
Δ = 102-4·1·(-19)
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{11}}{2*1}=\frac{-10-4\sqrt{11}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{11}}{2*1}=\frac{-10+4\sqrt{11}}{2}
$