15x2+20x-3=0

Solution for 15x2+20x-3=0 equation:

15x^2+20x-3=0

a = 15; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·15·(-3)
Δ = 580
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{145}}{2*15}=\frac{-20-2\sqrt{145}}{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{145}}{2*15}=\frac{-20+2\sqrt{145}}{30}
$