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0.62=5t^2
We move all terms to the left:
0.62-(5t^2)=0
a = -5; b = 0; c = +0.62;
Δ = b2-4ac
Δ = 02-4·(-5)·0.62
Δ = 12.4
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{12.4}}{2*-5}=\frac{0-\sqrt{12.4}}{-10} =-\frac{\sqrt{}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{12.4}}{2*-5}=\frac{0+\sqrt{12.4}}{-10} =\frac{\sqrt{}}{-10} $
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