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4t^2-45t=0
a = 4; b = -45; c = 0;
Δ = b2-4ac
Δ = -452-4·4·0
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-45}{2*4}=\frac{0}{8} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+45}{2*4}=\frac{90}{8} =11+1/4 $
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