20x^2+25x-12.5=0

Solution for 20x^2+25x-12.5=0 equation:

20x^2+25x-12.5=0

a = 20; b = 25; c = -12.5;
Δ = b2-4ac
Δ = 252-4·20·(-12.5)
Δ = 1625
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:

$\sqrt{\Delta}=\sqrt{1625}=\sqrt{25*65}=\sqrt{25}*\sqrt{65}=5\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{65}}{2*20}=\frac{-25-5\sqrt{65}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{65}}{2*20}=\frac{-25+5\sqrt{65}}{40}
$