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4t^2+44t-116=0
a = 4; b = 44; c = -116;
Δ = b2-4ac
Δ = 442-4·4·(-116)
Δ = 3792
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3792}=\sqrt{16*237}=\sqrt{16}*\sqrt{237}=4\sqrt{237}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{237}}{2*4}=\frac{-44-4\sqrt{237}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{237}}{2*4}=\frac{-44+4\sqrt{237}}{8} $
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