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5t^2-30t+25=0
a = 5; b = -30; c = +25;
Δ = b2-4ac
Δ = -302-4·5·25
Δ = 400
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{400}=20$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-20}{2*5}=\frac{10}{10} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+20}{2*5}=\frac{50}{10} =5 $
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