0.5d^2+d-14.4=0

Solution for 0.5d^2+d-14.4=0 equation:

0.5d^2+d-14.4=0

a = 0.5; b = 1; c = -14.4;
Δ = b2-4ac
Δ = 12-4·0.5·(-14.4)
Δ = 29.8
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{-(1)-\sqrt{29.8}}{2*0.5}=\frac{-1-\sqrt{29.8}}{1}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{29.8}}{2*0.5}=\frac{-1+\sqrt{29.8}}{1}
$