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t^2-4t=60
We move all terms to the left:
t^2-4t-(60)=0
a = 1; b = -4; c = -60;
Δ = b2-4ac
Δ = -42-4·1·(-60)
Δ = 256
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{256}=16$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*1}=\frac{-12}{2} =-6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*1}=\frac{20}{2} =10 $
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