r^2+10r-33=8

Solution for r^2+10r-33=8 equation:

r^2+10r-33=8

We move all terms to the left:

r^2+10r-33-(8)=0

We add all the numbers together, and all the variables

r^2+10r-41=0

a = 1; b = 10; c = -41;
Δ = b2-4ac
Δ = 102-4·1·(-41)
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{66}}{2*1}=\frac{-10-2\sqrt{66}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{66}}{2*1}=\frac{-10+2\sqrt{66}}{2}
$