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.625t^2-10=0
a = .625; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·.625·(-10)
Δ = 25
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-5}{2*.625}=\frac{-5}{1.25} =-4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+5}{2*.625}=\frac{5}{1.25} =4 $
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