(d2)=(3d+40)

Solution for (d2)=(3d+40) equation:

(d2)=(3d+40)

We move all terms to the left:

(d2)-((3d+40))=0

We add all the numbers together, and all the variables

d^2-((3d+40))=0


We calculate terms in parentheses: -((3d+40)), so:

(3d+40)

We get rid of parentheses

3d+40

Back to the equation:

-(3d+40)


We get rid of parentheses

d^2-3d-40=0

a = 1; b = -3; c = -40;
Δ = b2-4ac
Δ = -32-4·1·(-40)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*1}=\frac{-10}{2}
=-5
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*1}=\frac{16}{2}
=8
$