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-5t^2+40t-4=0
a = -5; b = 40; c = -4;
Δ = b2-4ac
Δ = 402-4·(-5)·(-4)
Δ = 1520
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{95}}{2*-5}=\frac{-40-4\sqrt{95}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{95}}{2*-5}=\frac{-40+4\sqrt{95}}{-10} $
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