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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+x^2+5x+4x+20))+30=0


We calculate terms in parentheses: -((+x^2+5x+4x+20)), so:

(+x^2+5x+4x+20)

We get rid of parentheses

x^2+5x+4x+20

We add all the numbers together, and all the variables

x^2+9x+20

Back to the equation:

-(x^2+9x+20)


We get rid of parentheses

-x^2-9x-20+30=0

We add all the numbers together, and all the variables

-1x^2-9x+10=0

a = -1; b = -9; c = +10;
Δ = b2-4ac
Δ = -92-4·(-1)·10
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-11}{2*-1}=\frac{-2}{-2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+11}{2*-1}=\frac{20}{-2}
=-10
$