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5t^2+7t=24
We move all terms to the left:
5t^2+7t-(24)=0
a = 5; b = 7; c = -24;
Δ = b2-4ac
Δ = 72-4·5·(-24)
Δ = 529
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{529}=23$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*5}=\frac{-30}{10} =-3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*5}=\frac{16}{10} =1+3/5 $
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