10x(2x+5)=20

Solution for 10x(2x+5)=20 equation:

10x(2x+5)=20

We move all terms to the left:

10x(2x+5)-(20)=0

We multiply parentheses

20x^2+50x-20=0

a = 20; b = 50; c = -20;
Δ = b2-4ac
Δ = 502-4·20·(-20)
Δ = 4100
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:

$\sqrt{\Delta}=\sqrt{4100}=\sqrt{100*41}=\sqrt{100}*\sqrt{41}=10\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{41}}{2*20}=\frac{-50-10\sqrt{41}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{41}}{2*20}=\frac{-50+10\sqrt{41}}{40}
$