5r^2+25r+30=0

Solution for 5r^2+25r+30=0 equation:

5r^2+25r+30=0

a = 5; b = 25; c = +30;
Δ = b2-4ac
Δ = 252-4·5·30
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5}{2*5}=\frac{-30}{10}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5}{2*5}=\frac{-20}{10}
=-2
$