20t(t+2)=17t-6

Solution for 20t(t+2)=17t-6 equation:

20t(t+2)=17t-6

We move all terms to the left:

20t(t+2)-(17t-6)=0

We multiply parentheses

20t^2+40t-(17t-6)=0

We get rid of parentheses

20t^2+40t-17t+6=0

We add all the numbers together, and all the variables

20t^2+23t+6=0

a = 20; b = 23; c = +6;
Δ = b2-4ac
Δ = 232-4·20·6
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*20}=\frac{-30}{40}
=-3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*20}=\frac{-16}{40}
=-2/5
$