4t^2+8t-32=0

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Solution for 4t^2+8t-32=0 equation: 



4t^2+8t-32=0

a = 4; b = 8; c = -32;
Δ = b2-4ac
Δ = 82-4·4·(-32)
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*4}=\frac{-32}{8}
=-4
$


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

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