4t^2+2t-6=0

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Solution for 4t^2+2t-6=0 equation: 



4t^2+2t-6=0

a = 4; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·4·(-6)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*4}=\frac{-12}{8}
=-1+1/2
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*4}=\frac{8}{8}
=1
$

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