t^2+6t-40=0

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Solution for t^2+6t-40=0 equation: 



t^2+6t-40=0

a = 1; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·1·(-40)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*1}=\frac{-20}{2}
=-10
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*1}=\frac{8}{2}
=4
$

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