r2+3r-54=0

Solution for r2+3r-54=0 equation:

r2+3r-54=0

We add all the numbers together, and all the variables

r^2+3r-54=0

a = 1; b = 3; c = -54;
Δ = b2-4ac
Δ = 32-4·1·(-54)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-15}{2*1}=\frac{-18}{2}
=-9
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+15}{2*1}=\frac{12}{2}
=6
$