((x+4)*(2x))/2=21

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

((x+4)(2x))/2=21

We move all terms to the left:

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

We multiply all the terms by the denominator


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


We calculate terms in parentheses: +((x+4)2x), so:

(x+4)2x

We multiply parentheses

2x^2+8x

Back to the equation:

+(2x^2+8x)


We add all the numbers together, and all the variables

(2x^2+8x)-42=0

We get rid of parentheses

2x^2+8x-42=0

a = 2; b = 8; c = -42;
Δ = b2-4ac
Δ = 82-4·2·(-42)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-20}{2*2}=\frac{-28}{4}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+20}{2*2}=\frac{12}{4}
=3
$