15b^2+14b-4=0

Solution for 15b^2+14b-4=0 equation:

15b^2+14b-4=0

a = 15; b = 14; c = -4;
Δ = b2-4ac
Δ = 142-4·15·(-4)
Δ = 436
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{109}}{2*15}=\frac{-14-2\sqrt{109}}{30}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{109}}{2*15}=\frac{-14+2\sqrt{109}}{30}
$