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5t^2+20t=585
We move all terms to the left:
5t^2+20t-(585)=0
a = 5; b = 20; c = -585;
Δ = b2-4ac
Δ = 202-4·5·(-585)
Δ = 12100
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{12100}=110$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-110}{2*5}=\frac{-130}{10} =-13 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+110}{2*5}=\frac{90}{10} =9 $
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