10r^2+22r+4=0

Solution for 10r^2+22r+4=0 equation:

10r^2+22r+4=0

a = 10; b = 22; c = +4;
Δ = b2-4ac
Δ = 222-4·10·4
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-18}{2*10}=\frac{-40}{20}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+18}{2*10}=\frac{-4}{20}
=-1/5
$