(4x^2)+(10x)+1=0

Solution for (4x^2)+(10x)+1=0 equation:

(4x^2)+(10x)+1=0

a = 4; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·4·1
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{21}}{2*4}=\frac{-10-2\sqrt{21}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{21}}{2*4}=\frac{-10+2\sqrt{21}}{8}
$