20=10x(2x+5)

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

20=10x(2x+5)

We move all terms to the left:

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


We calculate terms in parentheses: -(10x(2x+5)), so:

10x(2x+5)

We multiply parentheses

20x^2+50x

Back to the equation:

-(20x^2+50x)


We get rid of 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}
$