8r^2+22r+15=0

Solution for 8r^2+22r+15=0 equation:

8r^2+22r+15=0

a = 8; b = 22; c = +15;
Δ = b2-4ac
Δ = 222-4·8·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*8}=\frac{-24}{16}
=-1+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*8}=\frac{-20}{16}
=-1+1/4
$