10x^2+42x+44=0

Solution for 10x^2+42x+44=0 equation:

10x^2+42x+44=0

a = 10; b = 42; c = +44;
Δ = b2-4ac
Δ = 422-4·10·44
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2}{2*10}=\frac{-44}{20}
=-2+1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2}{2*10}=\frac{-40}{20}
=-2
$