r^2+10r-41=-10

Solution for r^2+10r-41=-10 equation:

r^2+10r-41=-10

We move all terms to the left:

r^2+10r-41-(-10)=0

We add all the numbers together, and all the variables

r^2+10r-31=0

a = 1; b = 10; c = -31;
Δ = b2-4ac
Δ = 102-4·1·(-31)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{14}}{2*1}=\frac{-10-4\sqrt{14}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{14}}{2*1}=\frac{-10+4\sqrt{14}}{2}
$