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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 $
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