(d+3)(d+3)=10(d+3)

Solution for (d+3)(d+3)=10(d+3) equation:

(d+3)(d+3)=10(d+3)

We move all terms to the left:

(d+3)(d+3)-(10(d+3))=0

We multiply parentheses ..

(+d^2+3d+3d+9)-(10(d+3))=0


We calculate terms in parentheses: -(10(d+3)), so:

10(d+3)

We multiply parentheses

10d+30

Back to the equation:

-(10d+30)


We get rid of parentheses

d^2+3d+3d-10d+9-30=0

We add all the numbers together, and all the variables

d^2-4d-21=0

a = 1; b = -4; c = -21;
Δ = b2-4ac
Δ = -42-4·1·(-21)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*1}=\frac{-6}{2}
=-3
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*1}=\frac{14}{2}
=7
$