10(2c+5)c-(1)=0

Solution for 10(2c+5)c-(1)=0 equation:

10(2c+5)c-(1)=0

We multiply parentheses

20c^2+50c-1=0

a = 20; b = 50; c = -1;
Δ = b2-4ac
Δ = 502-4·20·(-1)
Δ = 2580
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2580}=\sqrt{4*645}=\sqrt{4}*\sqrt{645}=2\sqrt{645}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{645}}{2*20}=\frac{-50-2\sqrt{645}}{40}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{645}}{2*20}=\frac{-50+2\sqrt{645}}{40}
$