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18=4+31t+-16t^2
We move all terms to the left:
18-(4+31t+-16t^2)=0
We use the square of the difference formula
-(4+31t-16t^2)+18=0
We get rid of parentheses
16t^2-31t-4+18=0
We add all the numbers together, and all the variables
16t^2-31t+14=0
a = 16; b = -31; c = +14;
Δ = b2-4ac
Δ = -312-4·16·14
Δ = 65
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{65}}{2*16}=\frac{31-\sqrt{65}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{65}}{2*16}=\frac{31+\sqrt{65}}{32} $
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