2.1x^2+0.6x-4=0

Solution for 2.1x^2+0.6x-4=0 equation:

2.1x^2+0.6x-4=0

a = 2.1; b = 0.6; c = -4;
Δ = b2-4ac
Δ = 0.62-4·2.1·(-4)
Δ = 33.96
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{-(0.6)-\sqrt{33.96}}{2*2.1}=\frac{-0.6-\sqrt{33.96}}{4.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.6)+\sqrt{33.96}}{2*2.1}=\frac{-0.6+\sqrt{33.96}}{4.2}
$