r^2-5r-30=27

Solution for r^2-5r-30=27 equation:

r^2-5r-30=27

We move all terms to the left:

r^2-5r-30-(27)=0

We add all the numbers together, and all the variables

r^2-5r-57=0

a = 1; b = -5; c = -57;
Δ = b2-4ac
Δ = -52-4·1·(-57)
Δ = 253
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{253}}{2*1}=\frac{5-\sqrt{253}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{253}}{2*1}=\frac{5+\sqrt{253}}{2}
$