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-5t^2+5t+30=0
a = -5; b = 5; c = +30;
Δ = b2-4ac
Δ = 52-4·(-5)·30
Δ = 625
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{625}=25$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-25}{2*-5}=\frac{-30}{-10} =+3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+25}{2*-5}=\frac{20}{-10} =-2 $
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