(d+5)(d-7)=d-5

Solution for (d+5)(d-7)=d-5 equation:

(d+5)(d-7)=d-5

We move all terms to the left:

(d+5)(d-7)-(d-5)=0

We get rid of parentheses

(d+5)(d-7)-d+5=0

We multiply parentheses ..

(+d^2-7d+5d-35)-d+5=0

We add all the numbers together, and all the variables

(+d^2-7d+5d-35)-1d+5=0

We get rid of parentheses

d^2-7d+5d-1d-35+5=0

We add all the numbers together, and all the variables

d^2-3d-30=0

a = 1; b = -3; c = -30;
Δ = b2-4ac
Δ = -32-4·1·(-30)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{129}}{2*1}=\frac{3-\sqrt{129}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*1}=\frac{3+\sqrt{129}}{2}
$