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5t^2-7t-30=0
a = 5; b = -7; c = -30;
Δ = b2-4ac
Δ = -72-4·5·(-30)
Δ = 649
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{649}}{2*5}=\frac{7-\sqrt{649}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{649}}{2*5}=\frac{7+\sqrt{649}}{10} $
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