(t+3)(t-3)=(t+3)2

Solution for (t+3)(t-3)=(t+3)2 equation:

(t+3)(t-3)=(t+3)2

We move all terms to the left:

(t+3)(t-3)-((t+3)2)=0

We use the square of the difference formula

t^2-((t+3)2)-9=0


We calculate terms in parentheses: -((t+3)2), so:

(t+3)2

We multiply parentheses

2t+6

Back to the equation:

-(2t+6)


We get rid of parentheses

t^2-2t-6-9=0

We add all the numbers together, and all the variables

t^2-2t-15=0

a = 1; b = -2; c = -15;
Δ = b2-4ac
Δ = -22-4·1·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*1}=\frac{-6}{2}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*1}=\frac{10}{2}
=5
$