(4+x)(5+x)-20=10

Solution for (4+x)(5+x)-20=10 equation:

(4+x)(5+x)-20=10

We move all terms to the left:

(4+x)(5+x)-20-(10)=0

We add all the numbers together, and all the variables

(x+4)(x+5)-20-10=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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