30=(2x+5)(2x+4)

Solution for 30=(2x+5)(2x+4) equation:

30=(2x+5)(2x+4)

We move all terms to the left:

30-((2x+5)(2x+4))=0

We multiply parentheses ..

-((+4x^2+8x+10x+20))+30=0


We calculate terms in parentheses: -((+4x^2+8x+10x+20)), so:

(+4x^2+8x+10x+20)

We get rid of parentheses

4x^2+8x+10x+20

We add all the numbers together, and all the variables

4x^2+18x+20

Back to the equation:

-(4x^2+18x+20)


We get rid of parentheses

-4x^2-18x-20+30=0

We add all the numbers together, and all the variables

-4x^2-18x+10=0

a = -4; b = -18; c = +10;
Δ = b2-4ac
Δ = -182-4·(-4)·10
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*-4}=\frac{-4}{-8}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*-4}=\frac{40}{-8}
=-5
$