3+4x-0.3x^2=0

Solution for 3+4x-0.3x^2=0 equation:

3+4x-0.3x^2=0

a = -0.3; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·(-0.3)·3
Δ = 19.6
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{-(4)-\sqrt{19.6}}{2*-0.3}=\frac{-4-\sqrt{19.6}}{-0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{19.6}}{2*-0.3}=\frac{-4+\sqrt{19.6}}{-0.6}
$