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t^2-5t=5
We move all terms to the left:
t^2-5t-(5)=0
a = 1; b = -5; c = -5;
Δ = b2-4ac
Δ = -52-4·1·(-5)
Δ = 45
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3\sqrt{5}}{2*1}=\frac{5-3\sqrt{5}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3\sqrt{5}}{2*1}=\frac{5+3\sqrt{5}}{2} $
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