20x2+15x-10=0

Solution for 20x2+15x-10=0 equation:

20x^2+15x-10=0

a = 20; b = 15; c = -10;
Δ = b2-4ac
Δ = 152-4·20·(-10)
Δ = 1025
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:

$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{41}}{2*20}=\frac{-15-5\sqrt{41}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{41}}{2*20}=\frac{-15+5\sqrt{41}}{40}
$