6t(t+4)=t-20

Solution for 6t(t+4)=t-20 equation:

6t(t+4)=t-20

We move all terms to the left:

6t(t+4)-(t-20)=0

We multiply parentheses

6t^2+24t-(t-20)=0

We get rid of parentheses

6t^2+24t-t+20=0

We add all the numbers together, and all the variables

6t^2+23t+20=0

a = 6; b = 23; c = +20;
Δ = b2-4ac
Δ = 232-4·6·20
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*6}=\frac{-30}{12}
=-2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*6}=\frac{-16}{12}
=-1+1/3
$