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5t^2+23t=0
a = 5; b = 23; c = 0;
Δ = b2-4ac
Δ = 232-4·5·0
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-23}{2*5}=\frac{-46}{10} =-4+3/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+23}{2*5}=\frac{0}{10} =0 $
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