D=5;(2d-1)(d+4)

Solution for D=5;(2d-1)(d+4) equation:

=5(2D-1)(D+4)

We move all terms to the left:

-(5(2D-1)(D+4))=0

We multiply parentheses ..

-(5(+2D^2+8D-1D-4))=0


We calculate terms in parentheses: -(5(+2D^2+8D-1D-4)), so:

5(+2D^2+8D-1D-4)

We multiply parentheses

10D^2+40D-5D-20

We add all the numbers together, and all the variables

10D^2+35D-20

Back to the equation:

-(10D^2+35D-20)


We get rid of parentheses

-10D^2-35D+20=0

a = -10; b = -35; c = +20;
Δ = b2-4ac
Δ = -352-4·(-10)·20
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-45}{2*-10}=\frac{-10}{-20}
=1/2
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+45}{2*-10}=\frac{80}{-20}
=-4
$