10b^2+19b+7=0

Solution for 10b^2+19b+7=0 equation:

10b^2+19b+7=0

a = 10; b = 19; c = +7;
Δ = b2-4ac
Δ = 192-4·10·7
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-9}{2*10}=\frac{-28}{20}
=-1+2/5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+9}{2*10}=\frac{-10}{20}
=-1/2
$