16t^2+80t+64=0

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Solution for 16t^2+80t+64=0 equation: 



16t^2+80t+64=0

a = 16; b = 80; c = +64;
Δ = b2-4ac
Δ = 802-4·16·64
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-48}{2*16}=\frac{-128}{32}
=-4
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+48}{2*16}=\frac{-32}{32}
=-1
$

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