3.2x+2x^2-5=0

Solution for 3.2x+2x^2-5=0 equation:

3.2x+2x^2-5=0

a = 2; b = 3.2; c = -5;
Δ = b2-4ac
Δ = 3.22-4·2·(-5)
Δ = 50.24
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{-(3.2)-\sqrt{50.24}}{2*2}=\frac{-3.2-\sqrt{50.24}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.2)+\sqrt{50.24}}{2*2}=\frac{-3.2+\sqrt{50.24}}{4}
$