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5t^2-30t-35=0
a = 5; b = -30; c = -35;
Δ = b2-4ac
Δ = -302-4·5·(-35)
Δ = 1600
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{1600}=40$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-40}{2*5}=\frac{-10}{10} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+40}{2*5}=\frac{70}{10} =7 $
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