35b^2+19b-24=0

Solution for 35b^2+19b-24=0 equation:

35b^2+19b-24=0

a = 35; b = 19; c = -24;
Δ = b2-4ac
Δ = 192-4·35·(-24)
Δ = 3721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3721}=61$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-61}{2*35}=\frac{-80}{70}
=-1+1/7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+61}{2*35}=\frac{42}{70}
=3/5
$