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4t^2+20t-280=0
a = 4; b = 20; c = -280;
Δ = b2-4ac
Δ = 202-4·4·(-280)
Δ = 4880
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4880}=\sqrt{16*305}=\sqrt{16}*\sqrt{305}=4\sqrt{305}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{305}}{2*4}=\frac{-20-4\sqrt{305}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{305}}{2*4}=\frac{-20+4\sqrt{305}}{8} $
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