2t2+8t-64=0

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Solution for 2t2+8t-64=0 equation: 



2t^2+8t-64=0

a = 2; b = 8; c = -64;
Δ = b2-4ac
Δ = 82-4·2·(-64)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-24}{2*2}=\frac{-32}{4}
=-8
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+24}{2*2}=\frac{16}{4}
=4
$

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