20=4(x+3)x=

Solution for 20=4(x+3)x= equation:

20=4(x+3)x=

We move all terms to the left:

20-(4(x+3)x)=0


We calculate terms in parentheses: -(4(x+3)x), so:

4(x+3)x

We multiply parentheses

4x^2+12x

Back to the equation:

-(4x^2+12x)


We get rid of parentheses

-4x^2-12x+20=0

a = -4; b = -12; c = +20;
Δ = b2-4ac
Δ = -122-4·(-4)·20
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{29}}{2*-4}=\frac{12-4\sqrt{29}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{29}}{2*-4}=\frac{12+4\sqrt{29}}{-8}
$