x(2x+4)=21

Solution for x(2x+4)=21 equation:

x(2x+4)=21

We move all terms to the left:

x(2x+4)-(21)=0

We multiply parentheses

2x^2+4x-21=0

a = 2; b = 4; c = -21;
Δ = b2-4ac
Δ = 42-4·2·(-21)
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{46}}{2*2}=\frac{-4-2\sqrt{46}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{46}}{2*2}=\frac{-4+2\sqrt{46}}{4}
$