r^2+8r-38=0

Solution for r^2+8r-38=0 equation:

r^2+8r-38=0

a = 1; b = 8; c = -38;
Δ = b2-4ac
Δ = 82-4·1·(-38)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-6\sqrt{6}}{2*1}=\frac{-8-6\sqrt{6}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+6\sqrt{6}}{2*1}=\frac{-8+6\sqrt{6}}{2}
$