-2t^2-20t-30=0

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Solution for -2t^2-20t-30=0 equation:



-2t^2-20t-30=0
a = -2; b = -20; c = -30;
Δ = b2-4ac
Δ = -202-4·(-2)·(-30)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{10}}{2*-2}=\frac{20-4\sqrt{10}}{-4} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{10}}{2*-2}=\frac{20+4\sqrt{10}}{-4} $

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