3x^2+2.5x-0.5=0

Solution for 3x^2+2.5x-0.5=0 equation:

3x^2+2.5x-0.5=0

a = 3; b = 2.5; c = -0.5;
Δ = b2-4ac
Δ = 2.52-4·3·(-0.5)
Δ = 12.25
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{-(2.5)-\sqrt{12.25}}{2*3}=\frac{-2.5-\sqrt{12.25}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.5)+\sqrt{12.25}}{2*3}=\frac{-2.5+\sqrt{12.25}}{6}
$