0=-16t^2+50t-15

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Solution for 0=-16t^2+50t-15 equation:



0=-16t^2+50t-15
We move all terms to the left:
0-(-16t^2+50t-15)=0
We add all the numbers together, and all the variables
-(-16t^2+50t-15)=0
We get rid of parentheses
16t^2-50t+15=0
a = 16; b = -50; c = +15;
Δ = b2-4ac
Δ = -502-4·16·15
Δ = 1540
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:
$\sqrt{\Delta}=\sqrt{1540}=\sqrt{4*385}=\sqrt{4}*\sqrt{385}=2\sqrt{385}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{385}}{2*16}=\frac{50-2\sqrt{385}}{32} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{385}}{2*16}=\frac{50+2\sqrt{385}}{32} $

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