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5t^2+15-20t=0
a = 5; b = -20; c = +15;
Δ = b2-4ac
Δ = -202-4·5·15
Δ = 100
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{100}=10$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10}{2*5}=\frac{10}{10} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10}{2*5}=\frac{30}{10} =3 $
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