15=-(d-5)(2d+6)

Solution for 15=-(d-5)(2d+6) equation:

15=-(d-5)(2d+6)

We move all terms to the left:

15-(-(d-5)(2d+6))=0

We multiply parentheses ..

-(-(+2d^2+6d-10d-30))+15=0


We calculate terms in parentheses: -(-(+2d^2+6d-10d-30)), so:

-(+2d^2+6d-10d-30)

We get rid of parentheses

-2d^2-6d+10d+30

We add all the numbers together, and all the variables

-2d^2+4d+30

Back to the equation:

-(-2d^2+4d+30)


We get rid of parentheses

2d^2-4d-30+15=0

We add all the numbers together, and all the variables

2d^2-4d-15=0

a = 2; b = -4; c = -15;
Δ = b2-4ac
Δ = -42-4·2·(-15)
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{34}}{2*2}=\frac{4-2\sqrt{34}}{4}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{34}}{2*2}=\frac{4+2\sqrt{34}}{4}
$