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100=5t^2
We move all terms to the left:
100-(5t^2)=0
a = -5; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-5)·100
Δ = 2000
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{5}}{2*-5}=\frac{0-20\sqrt{5}}{-10} =-\frac{20\sqrt{5}}{-10} =-\frac{2\sqrt{5}}{-1} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{5}}{2*-5}=\frac{0+20\sqrt{5}}{-10} =\frac{20\sqrt{5}}{-10} =\frac{2\sqrt{5}}{-1} $
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