x^2+4.2x+1.26=0

Solution for x^2+4.2x+1.26=0 equation:

x^2+4.2x+1.26=0

a = 1; b = 4.2; c = +1.26;
Δ = b2-4ac
Δ = 4.22-4·1·1.26
Δ = 12.6
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{-(4.2)-\sqrt{12.6}}{2*1}=\frac{-4.2-\sqrt{12.6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.2)+\sqrt{12.6}}{2*1}=\frac{-4.2+\sqrt{12.6}}{2}
$