4(2x)+22x=(x+2)(x+3)

Solution for 4(2x)+22x=(x+2)(x+3) equation:

4(2x)+22x=(x+2)(x+3)

We move all terms to the left:

4(2x)+22x-((x+2)(x+3))=0

We add all the numbers together, and all the variables

64x-((x+2)(x+3))=0

We multiply parentheses ..

-((+x^2+3x+2x+6))+64x=0


We calculate terms in parentheses: -((+x^2+3x+2x+6)), so:

(+x^2+3x+2x+6)

We get rid of parentheses

x^2+3x+2x+6

We add all the numbers together, and all the variables

x^2+5x+6

Back to the equation:

-(x^2+5x+6)


We add all the numbers together, and all the variables

64x-(x^2+5x+6)=0

We get rid of parentheses

-x^2+64x-5x-6=0

We add all the numbers together, and all the variables

-1x^2+59x-6=0

a = -1; b = 59; c = -6;
Δ = b2-4ac
Δ = 592-4·(-1)·(-6)
Δ = 3457
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(59)-\sqrt{3457}}{2*-1}=\frac{-59-\sqrt{3457}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(59)+\sqrt{3457}}{2*-1}=\frac{-59+\sqrt{3457}}{-2}
$