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-17.5t^2+30t+6=0
a = -17.5; b = 30; c = +6;
Δ = b2-4ac
Δ = 302-4·(-17.5)·6
Δ = 1320
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1320}=\sqrt{4*330}=\sqrt{4}*\sqrt{330}=2\sqrt{330}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{330}}{2*-17.5}=\frac{-30-2\sqrt{330}}{-35} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{330}}{2*-17.5}=\frac{-30+2\sqrt{330}}{-35} $
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