(3t-2)(t+5)=(t-3)(t-4)

Solution for (3t-2)(t+5)=(t-3)(t-4) equation:

(3t-2)(t+5)=(t-3)(t-4)

We move all terms to the left:

(3t-2)(t+5)-((t-3)(t-4))=0

We multiply parentheses ..

(+3t^2+15t-2t-10)-((t-3)(t-4))=0


We calculate terms in parentheses: -((t-3)(t-4)), so:

(t-3)(t-4)

We multiply parentheses ..

(+t^2-4t-3t+12)

We get rid of parentheses

t^2-4t-3t+12

We add all the numbers together, and all the variables

t^2-7t+12

Back to the equation:

-(t^2-7t+12)


We get rid of parentheses

3t^2-t^2+15t-2t+7t-10-12=0

We add all the numbers together, and all the variables

2t^2+20t-22=0

a = 2; b = 20; c = -22;
Δ = b2-4ac
Δ = 202-4·2·(-22)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-24}{2*2}=\frac{-44}{4}
=-11
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+24}{2*2}=\frac{4}{4}
=1
$