20t(t+2)=9t-12

Solution for 20t(t+2)=9t-12 equation:

20t(t+2)=9t-12

We move all terms to the left:

20t(t+2)-(9t-12)=0

We multiply parentheses

20t^2+40t-(9t-12)=0

We get rid of parentheses

20t^2+40t-9t+12=0

We add all the numbers together, and all the variables

20t^2+31t+12=0

a = 20; b = 31; c = +12;
Δ = b2-4ac
Δ = 312-4·20·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-1}{2*20}=\frac{-32}{40}
=-4/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+1}{2*20}=\frac{-30}{40}
=-3/4
$