(1+d)(d+2)=-(9d-26)

Solution for (1+d)(d+2)=-(9d-26) equation:

(1+d)(d+2)=-(9d-26)

We move all terms to the left:

(1+d)(d+2)-(-(9d-26))=0

We add all the numbers together, and all the variables

(d+1)(d+2)-(-(9d-26))=0

We multiply parentheses ..

(+d^2+2d+d+2)-(-(9d-26))=0


We calculate terms in parentheses: -(-(9d-26)), so:

-(9d-26)

We get rid of parentheses

-9d+26

Back to the equation:

-(-9d+26)


We get rid of parentheses

d^2+2d+d+9d+2-26=0

We add all the numbers together, and all the variables

d^2+12d-24=0

a = 1; b = 12; c = -24;
Δ = b2-4ac
Δ = 122-4·1·(-24)
Δ = 240
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{15}}{2*1}=\frac{-12-4\sqrt{15}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{15}}{2*1}=\frac{-12+4\sqrt{15}}{2}
$