21(6)=(4x+7)(4x+3)

Solution for 21(6)=(4x+7)(4x+3) equation:

21(6)=(4x+7)(4x+3)

We move all terms to the left:

21(6)-((4x+7)(4x+3))=0

We multiply parentheses ..

-((+16x^2+12x+28x+21))+216=0


We calculate terms in parentheses: -((+16x^2+12x+28x+21)), so:

(+16x^2+12x+28x+21)

We get rid of parentheses

16x^2+12x+28x+21

We add all the numbers together, and all the variables

16x^2+40x+21

Back to the equation:

-(16x^2+40x+21)


We get rid of parentheses

-16x^2-40x-21+216=0

We add all the numbers together, and all the variables

-16x^2-40x+195=0

a = -16; b = -40; c = +195;
Δ = b2-4ac
Δ = -402-4·(-16)·195
Δ = 14080
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:

$\sqrt{\Delta}=\sqrt{14080}=\sqrt{256*55}=\sqrt{256}*\sqrt{55}=16\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-16\sqrt{55}}{2*-16}=\frac{40-16\sqrt{55}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+16\sqrt{55}}{2*-16}=\frac{40+16\sqrt{55}}{-32}
$