0=(-5t-3)(t-6)

Solution for 0=(-5t-3)(t-6) equation:

0=(-5t-3)(t-6)

We move all terms to the left:

0-((-5t-3)(t-6))=0

We add all the numbers together, and all the variables

-((-5t-3)(t-6))=0

We multiply parentheses ..

-((-5t^2+30t-3t+18))=0


We calculate terms in parentheses: -((-5t^2+30t-3t+18)), so:

(-5t^2+30t-3t+18)

We get rid of parentheses

-5t^2+30t-3t+18

We add all the numbers together, and all the variables

-5t^2+27t+18

Back to the equation:

-(-5t^2+27t+18)


We get rid of parentheses

5t^2-27t-18=0

a = 5; b = -27; c = -18;
Δ = b2-4ac
Δ = -272-4·5·(-18)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-33}{2*5}=\frac{-6}{10}
=-3/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+33}{2*5}=\frac{60}{10}
=6
$