0.32x^2+x-1.25=0

Solution for 0.32x^2+x-1.25=0 equation:

0.32x^2+x-1.25=0

a = 0.32; b = 1; c = -1.25;
Δ = b2-4ac
Δ = 12-4·0.32·(-1.25)
Δ = 2.6
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{2.6}}{2*0.32}=\frac{-1-\sqrt{2.6}}{0.64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{2.6}}{2*0.32}=\frac{-1+\sqrt{2.6}}{0.64}
$