(2x+1)x=21

Solution for (2x+1)x=21 equation:

(2x+1)x=21

We move all terms to the left:

(2x+1)x-(21)=0

We multiply parentheses

2x^2+x-21=0

a = 2; b = 1; c = -21;
Δ = b2-4ac
Δ = 12-4·2·(-21)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*2}=\frac{-14}{4}
=-3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*2}=\frac{12}{4}
=3
$