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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

5t^2-26t-18=0

a = 5; b = -26; c = -18;
Δ = b2-4ac
Δ = -262-4·5·(-18)
Δ = 1036
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1036}=\sqrt{4*259}=\sqrt{4}*\sqrt{259}=2\sqrt{259}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{259}}{2*5}=\frac{26-2\sqrt{259}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{259}}{2*5}=\frac{26+2\sqrt{259}}{10}
$