((x+2)(x+5))-40=0

Solution for ((x+2)(x+5))-40=0 equation:

((x+2)(x+5))-40=0

We multiply parentheses ..

((+x^2+5x+2x+10))-40=0


We calculate terms in parentheses: +((+x^2+5x+2x+10)), so:

(+x^2+5x+2x+10)

We get rid of parentheses

x^2+5x+2x+10

We add all the numbers together, and all the variables

x^2+7x+10

Back to the equation:

+(x^2+7x+10)


We get rid of parentheses

x^2+7x+10-40=0

We add all the numbers together, and all the variables

x^2+7x-30=0

a = 1; b = 7; c = -30;
Δ = b2-4ac
Δ = 72-4·1·(-30)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*1}=\frac{-20}{2}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*1}=\frac{6}{2}
=3
$