r^2+r-38=0

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

r^2+r-38=0

a = 1; b = 1; c = -38;
Δ = b2-4ac
Δ = 12-4·1·(-38)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{17}}{2*1}=\frac{-1-3\sqrt{17}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{17}}{2*1}=\frac{-1+3\sqrt{17}}{2}
$