7+14t-16t^2=0

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Solution for 7+14t-16t^2=0 equation:



7+14t-16t^2=0
a = -16; b = 14; c = +7;
Δ = b2-4ac
Δ = 142-4·(-16)·7
Δ = 644
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:
$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{161}}{2*-16}=\frac{-14-2\sqrt{161}}{-32} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{161}}{2*-16}=\frac{-14+2\sqrt{161}}{-32} $

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