3(t+3)(t-1)=(t+2)(t+3)

Solution for 3(t+3)(t-1)=(t+2)(t+3) equation:

3(t+3)(t-1)=(t+2)(t+3)

We move all terms to the left:

3(t+3)(t-1)-((t+2)(t+3))=0

We multiply parentheses ..

3(+t^2-1t+3t-3)-((t+2)(t+3))=0


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

(t+2)(t+3)

We multiply parentheses ..

(+t^2+3t+2t+6)

We get rid of parentheses

t^2+3t+2t+6

We add all the numbers together, and all the variables

t^2+5t+6

Back to the equation:

-(t^2+5t+6)


We multiply parentheses

3t^2-3t+9t-(t^2+5t+6)-9=0

We get rid of parentheses

3t^2-t^2-3t+9t-5t-6-9=0

We add all the numbers together, and all the variables

2t^2+t-15=0

a = 2; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·2·(-15)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*2}=\frac{-12}{4}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*2}=\frac{10}{4}
=2+1/2
$