t^2+7t-120=0

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Solution for t^2+7t-120=0 equation:



t^2+7t-120=0
a = 1; b = 7; c = -120;
Δ = b2-4ac
Δ = 72-4·1·(-120)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*1}=\frac{-30}{2} =-15 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*1}=\frac{16}{2} =8 $

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