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2t^2-50t-315=0
a = 2; b = -50; c = -315;
Δ = b2-4ac
Δ = -502-4·2·(-315)
Δ = 5020
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5020}=\sqrt{4*1255}=\sqrt{4}*\sqrt{1255}=2\sqrt{1255}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{1255}}{2*2}=\frac{50-2\sqrt{1255}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{1255}}{2*2}=\frac{50+2\sqrt{1255}}{4} $
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