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15=2+30t-16t^2
We move all terms to the left:
15-(2+30t-16t^2)=0
We get rid of parentheses
16t^2-30t-2+15=0
We add all the numbers together, and all the variables
16t^2-30t+13=0
a = 16; b = -30; c = +13;
Δ = b2-4ac
Δ = -302-4·16·13
Δ = 68
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{17}}{2*16}=\frac{30-2\sqrt{17}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{17}}{2*16}=\frac{30+2\sqrt{17}}{32} $
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