10a^2+20a-1=0

Solution for 10a^2+20a-1=0 equation:

10a^2+20a-1=0

a = 10; b = 20; c = -1;
Δ = b2-4ac
Δ = 202-4·10·(-1)
Δ = 440
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{110}}{2*10}=\frac{-20-2\sqrt{110}}{20}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{110}}{2*10}=\frac{-20+2\sqrt{110}}{20}
$