d^2+-12d+-49=0

Solution for d^2+-12d+-49=0 equation:

d^2+-12d+-49=0

We add all the numbers together, and all the variables

d^2-12d=0

a = 1; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·1·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*1}=\frac{0}{2}
=0
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*1}=\frac{24}{2}
=12
$