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8t^2-45t=0
a = 8; b = -45; c = 0;
Δ = b2-4ac
Δ = -452-4·8·0
Δ = 2025
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{2025}=45$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-45}{2*8}=\frac{0}{16} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+45}{2*8}=\frac{90}{16} =5+5/8 $
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