10x^2+42x+28=0

Solution for 10x^2+42x+28=0 equation:

10x^2+42x+28=0

a = 10; b = 42; c = +28;
Δ = b2-4ac
Δ = 422-4·10·28
Δ = 644
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{161}}{2*10}=\frac{-42-2\sqrt{161}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{161}}{2*10}=\frac{-42+2\sqrt{161}}{20}
$