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0=4t^2+5t
We move all terms to the left:
0-(4t^2+5t)=0
We add all the numbers together, and all the variables
-(4t^2+5t)=0
We get rid of parentheses
-4t^2-5t=0
a = -4; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-4)·0
Δ = 25
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{25}=5$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-4}=\frac{0}{-8} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-4}=\frac{10}{-8} =-1+1/4 $
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