(x+2)(x+5)+3=(21-5)=9

Solution for (x+2)(x+5)+3=(21-5)=9 equation:

(x+2)(x+5)+3=(21-5)=9

We move all terms to the left:

(x+2)(x+5)+3-((21-5))=0

We add all the numbers together, and all the variables

(x+2)(x+5)+3-(16)=0

We add all the numbers together, and all the variables

(x+2)(x+5)-13=0

We multiply parentheses ..

(+x^2+5x+2x+10)-13=0

We get rid of parentheses

x^2+5x+2x+10-13=0

We add all the numbers together, and all the variables

x^2+7x-3=0

a = 1; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·1·(-3)
Δ = 61
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{-(7)-\sqrt{61}}{2*1}=\frac{-7-\sqrt{61}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{61}}{2*1}=\frac{-7+\sqrt{61}}{2}
$