0=-5(t^2-6t+4)

Solution for 0=-5(t^2-6t+4) equation:

0=-5(t^2-6t+4)

We move all terms to the left:

0-(-5(t^2-6t+4))=0

We add all the numbers together, and all the variables

-(-5(t^2-6t+4))=0


We calculate terms in parentheses: -(-5(t^2-6t+4)), so:

-5(t^2-6t+4)

We multiply parentheses

-5t^2+30t-20

Back to the equation:

-(-5t^2+30t-20)


We get rid of parentheses

5t^2-30t+20=0

a = 5; b = -30; c = +20;
Δ = b2-4ac
Δ = -302-4·5·20
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-10\sqrt{5}}{2*5}=\frac{30-10\sqrt{5}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+10\sqrt{5}}{2*5}=\frac{30+10\sqrt{5}}{10}
$