2x^2+3.5x=0

Solution for 2x^2+3.5x=0 equation:

2x^2+3.5x=0

a = 2; b = 3.5; c = 0;
Δ = b2-4ac
Δ = 3.52-4·2·0
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{-(3.5)-\sqrt{12.25}}{2*2}=\frac{-3.5-\sqrt{12.25}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{12.25}}{2*2}=\frac{-3.5+\sqrt{12.25}}{4}
$