35r^2+39r+10=0

Solution for 35r^2+39r+10=0 equation:

35r^2+39r+10=0

a = 35; b = 39; c = +10;
Δ = b2-4ac
Δ = 392-4·35·10
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-11}{2*35}=\frac{-50}{70}
=-5/7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+11}{2*35}=\frac{-28}{70}
=-2/5
$